# 3.2 - Quadratic equations and functions.

- Quadratic functions:

$3x^2 - 2x + 4$

$-4x^2 + 3x + 1$

$5x^2 - 2x$ (this is $5x^2 - 2x + 0$)

$-2x^2 + 4$ (this is $-2x^2 + 0x + 4$)

What about these? Yes or no?

$7x^2$

$\sqrt{3}x^2 - 2x + 4$

$2x$

$5$

- Quadratic function: $f(x) = ax^2 + bx + c$. $a \neq 0$.

- Quadratic equation: $ax^2 + bx + c = 0$. $a \neq 0$.

## Method of solution of $ax^2 + bx + c = 0$: Factoring.

- Try to write $ax^2 + bx + c$ as the product of linear factors by doing some guesswork.

- If $AB = 0$ then $A = 0$ or $B = 0$.

### Example 1. Solve $(3x-7)(5x+9)=0$

Think of $A=3x-7$ and $B=5x+9$.

| Step | Solve $A = 0$ | Solve $B = 0$ |
|-----------------------------|-----------------------------------|-------------------------------------|
| Set equation to zero | $3x - 7 = 0$ | $5x + 9 = 0$ |
| Isolate $x$ | $3x = 7$ | $5x = -9$ |
| Solve for $x$ | $x = \frac{7}{3}$ | $x = -\frac{9}{5}$ |

Solution: $x = \frac{7}{3}$ or $x = -\frac{9}{5}$.

---

### Example 2. Solve $x^2 - 36 = 0$

Write $x^2 - 36$ as a difference of squares: $x^2-36=(x+6)(x-6)$

Click on footnote for FOIL verification:[^1]

[^1]: \begin{align} (x+6)(x-6)&=(x)(x)+(x)(-6)+(6)(x)+(6)(-6) \\ &=x^2-6x+6x-36 \\
&=x^2-36
\end{align}

$$(x+6)(x-6)=0$$

Think of $A=x+6$ and $B=x-6$.

| Step | Solve $A=0$ | Solve $B=0$ |
|-----------------------------|--------------------------------------------|----|
| Set equation to zero | $x+6=0$ | $x-6=0$ |
| Isolate $x$ | $x=-6$ | $x=6$ |

Solution: $x = 6$ or $x = -6$.

- Typically get two solutions, unless the factors are the same.

---

### Example 3. $x^2+7x+12=0$

If $$x^2+7x+12=(x+A)(x+B)$$

\begin{align}
(x+A)(x+B)&=(x)(x)+(x)(B)+(A)(x)+(A)(B) \\
&=x^2+Bx+Ax+AB \\
&=x^2+(Ax+Bx)+(AB) \\
&=x^2+ \hspace{0.3cm} \quad 7x \hspace{0.9cm} \quad+12
\end{align}

then when we multiply out (FOIL) $AB$ must be $12$ and $Ax+Bx$ must be $7x$. So we seek $A,B$ so that $A+B=7$ and $AB=12$. Possibilities for $(A,B)$ multiplying to $12$ include pairs in this table:

| A | B | AB |
|---|----|---|
| 1 | 12 | 12 |
| 2 | 6 | 12 |
| 3 | 4 | 12|

Then add another column $A+B$ by adding the columns $A$ and $B$:

| A | B | A+B | AB |
|---|----|-----|----|
| 1 | 12 | 13 |12|
| 2 | 6 | 8 |12|
| **3** | **4** | **7** |**12**|

that $3+4=7$ and $(3)(4)=12$ so

$$x^2+7x+12=(x+3)(x+4)$$

Check the footnote for FOIL verification:[^2]

[^2]: \begin{align} (x+3)(x+4)&=(x)(x)+(x)(4)+(3)(x)+(3)(4) \\
&=x^2+4x+3x+12 \\
&=x^2+7x+12
\end{align}

Solve:

$$(x+3)(x+4)=0$$

\begin{align}
x+3&=0 & x+4&=0 \\
x&=-3 & x&=-4
\end{align}

- Guesswork is the basis of factoring - although there are some tricks to speed up the process.

---

### Example 4. $2x^2+4x-16=0$

- First notice $2$ divides every term. Factor it out:

$$2x^2+4x-16=2(x^2+2x-8)=2(x+A)(x+B)$$

Now we need $A,B$ so $AB=-8$ and $A+B=2$. Since the product is negative, one of $A,B$ is negative and one positive. So really we will have a difference of two numbers equal to $2$.

Possibilities for $(A,B)$ include:
$(1,-8)$, $(8,-1)$, $(-1,8)$, $(-8,1)$, $(4,-2)$,
$(-4,2)$, $(2,-4)$, $(-2,4)$, etc.

Since $4+(-2)=2$, $A=4$ and $B=-2$. So
$$2x^2+4x-16=2(x^2+2x-8)=2(x+4)(x-2)=0$$

\begin{align}
x+4&=0 & x-2&=0 \\
x&=-4 & x&=2
\end{align}

---

### Example 5. $3x^2-33=0$

We could write $$3x^2-33=3(x^2-11)=3(x+A)(x+B).$$
If we write $11$ as $(\sqrt{11})^2$, then we can factor as the difference of two squares.
But that is a little messy and there is another way of thinking of this:

- Because there is no $x$ term, we can isolate $x^2$ as in a linear equation:

$$3x^2=33$$
$$x^2=\frac{33}{3}=11$$

Here we use the `principle of square roots': we have $x=\sqrt{11}$ or $x=-\sqrt{11}$.
This is usually better than factoring.

- Usually write $x=\pm\sqrt{11}$

- Don't forget the solution can be plus or minus!

---

### Example 6. $7x^2+42=0$

Then $$7x^2=-42,$$ $$x^2=-\frac{42}{7},$$ $$x^2=-6.$$

If $x$ is real then $x^2$ cannot be negative, so for real numbers there is NO REAL SOLUTION.

- But usually we don't express it this way: we introduce the `imaginary' number
$i$ which satisfies $i^2=-1$.

- Then the solution is written $$x=\pm\sqrt{6}i$$

We don't emphasize imaginary numbers in this course, but since most books
talk a little about them, this is how answers are written. But it is still the case that there is no REAL solution.

---

### Example 7. $27x=9x^2$

- Always create an equation which is set equal to ZERO:

$$-9x^2+27x=0$$

- It's best if the $x^2$ has a positive coefficient. Multiply by $-1$:

$$9x^2-27x=0$$

Here both terms have $x$ in them. So factor it out, along with $9$:

$$9x(x-3)=0$$

Then $9x=0$ or $x-3=0$ so $x=\frac{0}{9}=0$ or $x=3$.

---

### Example 8. $x^2-4x-21=0$

#### Steps to Factor $x^2 - 4x - 21$

#### Step 1: Identify the constant term and the middle term
- The constant term is $-21$ (the number without $x$).
- The middle term is $-4x$ (the coefficient of $x$ is $-4$).

#### Step 2: Find two numbers that multiply to $-21$ and add to $-4$
We need to find two numbers that:
- **Multiply** to $-21$ (constant term).
- **Add** to $-4$ (middle term).

#### Step 3: List the factor pairs of $-21$ and check their sums

#### Factor Pairs of -21 and Their Sums

| Factor 1 | Factor 2 | Sum |
|----------|----------|------|
| 1 | -21 | -20 |
| -1 | 21 | 20 |
| 3 | -7 | **-4** |
| -3 | 7 | 4 |

#### Step 4: Select the correct factor pair
The correct factor pair is $3$ and $-7$, since:
- $3 \times -7 = -21$
- $3 + (-7) = -4$

#### Step 5: Write the factored form
Now that we have the correct factor pair, we can write the quadratic expression in its factored form:
$$
x^2 - 4x - 21 = (x + 3)(x - 7)
$$

#### Step 6: Solve for $x$
To solve the equation $(x+3)(x-7) = 0$, we set each factor to zero:
$$
x + 3 = 0 \quad \text{or} \quad x - 7 = 0
$$
Thus, the solutions are:
$$
x = -3 \quad \text{or} \quad x = 7
$$

#### Final Answer
The factored form of $x^2 - 4x - 21$ is $(x + 3)(x - 7)$, and the solutions are $x = -3$ and $x = 7$.

---

### Example 9. $x^2-8x+12=0$

#### Steps to Factor $x^2 - 8x + 12$

#### Step 1: Identify the constant term and the middle term
- The constant term is $12$ (the number without $x$).
- The middle term is $-8x$ (the coefficient of $x$ is $-8$).

#### Step 2: Find two numbers that multiply to $12$ and add to $-8$
We need to find two numbers that:
- **Multiply** to $12$ (constant term).
- **Add** to $-8$ (middle term).

#### Step 3: List the factor pairs of $12$ and check their sums

#### Factor Pairs of 12 and Their Sums

| Factor 1 | Factor 2 | Sum |
|----------|----------|------|
| 1 | 12 | 13 |
| -1 | -12 | -13 |
| 2 | 6 | 8 |
| **-2** | **-6** | **-8** |
| 3 | 4 | 7 |
| -3 | -4 | -7 |

#### Step 4: Select the correct factor pair
The correct factor pair is $-2$ and $-6$, since:
- $-2 \times -6 = 12$
- $-2 + (-6) = -8$

#### Step 5: Write the factored form
Now that we have the correct factor pair, we can write the quadratic expression in its factored form:
$$
x^2 - 8x + 12 = (x - 6)(x - 2)
$$

#### Step 6: Solve for $x$
To solve the equation $(x-6)(x-2) = 0$, we set each factor to zero:
$$
x - 6 = 0 \quad \text{or} \quad x - 2 = 0
$$
Thus, the solutions are:
$$
x = 6 \quad \text{or} \quad x = 2
$$

#### Final Answer
The factored form of $x^2 - 8x + 12$ is $(x - 6)(x - 2)$, and the solutions are $x = 6$ and $x = 2$.

---

### Example 10. $7x^2+22x+3=0$
$$(7x+\dots)(x+\dots)=0$$

#### Easiest Way to Factor $7x^2 + 22x + 3$

#### Step 1: Factor $7x^2$ and $3$
- $7x^2$ factors as $(7x)(x)$.
- $3$ can factor as $(1)(3)$, $(3)(1)$, $(-1)(-3)$, or $(-3)(-1)$.

#### Step 2: Check possible pairings

1. **First pairing**: $(7x + 1)(x + 3)$
Expand:
$$
(7x + 1)(x + 3) = 7x^2 + 22x + 3
$$
This matches the original equation.

2. **Second pairing**: $(7x + 3)(x + 1)$
Expand:
$$
(7x + 3)(x + 1) =7x^2 + 10x + 3
$$
This does not match.

3. **Third pairing**: $(7x - 1)(x - 3)$
Expand:
$$
(7x - 1)(x - 3) = 7x^2 - 22x + 3
$$
This doesn't match because it gives $-22x$, which is the opposite sign.

4. **Third pairing**: $(7x - 3)(x - 1)$
Expand:
$$
(7x - 3)(x - 1) = 7x^2 - 10x + 3
$$
This does not match.

#### Step 3: Conclusion
The correct factorization is:
$$
(7x + 1)(x + 3)
$$

#### Step 4: Solve for $x$
Set each factor to zero:
$$
7x + 1 = 0 \quad \text{or} \quad x + 3 = 0
$$
Thus, the solutions are:
$$
x = -\frac{1}{7} \quad \text{or} \quad x = -3
$$

#### Final Answer
The factored form of $7x^2 + 22x + 3$ is $(7x + 1)(x + 3)$, and the solutions are $x = -\frac{1}{7}$ and $x = -3$.

---

### Example 11. $y^3-5y^2-24y=0$

#### Steps to Factor $y^3 - 5y^2 - 24y$

#### Step 1: Factor out the greatest common factor
The greatest common factor (GCF) of all the terms is $y$, so we factor that out first:
$$
y^3 - 5y^2 - 24y = y(y^2 - 5y - 24)
$$

Now we need to factor $y^2 - 5y - 24$.

#### Step 2: Identify the constant term and the middle term
- The constant term is $-24$.
- The middle term is $-5y$.

#### Step 3: Find two numbers that multiply to $-24$ and add to $-5$
We need to find two numbers that:
- **Multiply** to $-24$ (constant term).
- **Add** to $-5$ (middle term).

#### Step 4: List the factor pairs of $-24$ and check their sums

#### Factor Pairs of -24 and Their Sums

| Factor 1 | Factor 2 | Sum |
|----------|----------|------|
| 1 | -24 | -23 |
| -1 | 24 | 23 |
| 2 | -12 | -10 |
| -2 | 12 | 10 |
| **3** | **-8** | **-5** |
| -3 | 8 | 5 |

#### Step 5: Select the correct factor pair
The correct factor pair is $3$ and $-8$, since:
- $3 \times -8 = -24$
- $3 + (-8) = -5$

#### Step 6: Write the factored form
Now that we have the correct factor pair, we can write the quadratic expression in its factored form:
$$
y^2 - 5y - 24 = (y + 3)(y - 8)
$$

Thus, the full factored form of the original expression is:
$$
y(y + 3)(y - 8) = 0
$$

#### Step 7: Solve for $y$
Set each factor to zero:
$$
y = 0 \quad \text{or} \quad y + 3 = 0 \quad \text{or} \quad y - 8 = 0
$$
Thus, the solutions are:
$$
y = 0 \quad \text{or} \quad y = -3 \quad \text{or} \quad y = 8
$$

#### Final Answer
The factored form of $y^3 - 5y^2 - 24y$ is $y(y + 3)(y - 8)$, and the solutions are $y = 0$, $y = -3$, and $y = 8$.

---

### Example 12. $y^4-8y^2+15=0$
$$(y^2+\dots)(y^2+\dots)=0$$

#### Steps to Factor $y^4 - 8y^2 + 15$

#### Step 1: Recognize the structure
This is a quadratic in terms of $y^2$. Let’s rewrite it as:
$$
(y^2)^2 - 8y^2 + 15 = 0
$$

Now we will factor it like a regular quadratic.

#### Step 2: Identify the constant term and the middle term
- The constant term is $15$.
- The middle term is $-8y^2$.

#### Step 3: Find two numbers that multiply to $15$ and add to $-8$
We need to find two numbers that:
- **Multiply** to $15$ (constant term).
- **Add** to $-8$ (middle term).

#### Step 4: List the factor pairs of $15$ and check their sums

#### Factor Pairs of 15 and Their Sums

| Factor 1 | Factor 2 | Sum |
|----------|----------|------|
| 1 | 15 | 16 |
| -1 | -15 | -16 |
| 3 | 5 | 8 |
| **-3** | **-5** | **-8** |

#### Step 5: Select the correct factor pair
The correct factor pair is $-3$ and $-5$, since:
- $-3 \times -5 = 15$
- $-3 + (-5) = -8$

#### Step 6: Write the factored form
Now that we have the correct factor pair, we can write the quadratic expression in its factored form:
$$
y^4 - 8y^2 + 15 = (y^2 - 3)(y^2 - 5)
$$

#### Step 7: Solve for $y^2$
We now set each factor to zero:
$$
y^2 - 3 = 0 \quad \text{or} \quad y^2 - 5 = 0
$$

Solving these gives:
$$
y^2 = 3 \quad \text{or} \quad y^2 = 5
$$

#### Step 8: Solve for $y$
Now, take the square root of both sides:
$$
y = \pm \sqrt{3} \quad \text{or} \quad y = \pm \sqrt{5}
$$

#### Final Answer
The factored form of $y^4 - 8y^2 + 15$ is $(y^2 - 3)(y^2 - 5)$, and the solutions are:
$$
y = \pm \sqrt{3} \quad \text{and} \quad y = \pm \sqrt{5}
$$

---

## Graph of $f(x)=ax^2+bx+c$.

- Is always a `parabola' - a U shape if $a>0$.


- Upside down U if $a<0$.


---

- `Zeroes' of $f(x)$ are the solutions to $ax^2+bx+c=0$.

Example. Find the $x$-intercepts of $f(x)=-x^2+8x-7=0$.

### Steps to Find the $x$-Intercepts

#### Step 1: Identify the equation and rearrange if necessary
We are given the equation:
$$
-x^2 + 8x - 7 = 0
$$
To make factoring easier, multiply both sides by $-1$ to get:
$$
x^2 - 8x + 7 = 0
$$

#### Step 2: Identify the constant term and the middle term
- The constant term is $7$.
- The middle term is $-8x$.

#### Step 3: Find two numbers that multiply to $7$ and add to $-8$
We need to find two numbers that:
- **Multiply** to $7$ (constant term).
- **Add** to $-8$ (middle term).

#### Step 4: List the factor pairs of $7$ and check their sums

#### Factor Pairs of 7 and Their Sums

| Factor 1 | Factor 2 | Sum |
|----------|----------|------|
| 1 | 7 | 8 |
| **-1** | **-7** | **-8** |

#### Step 5: Select the correct factor pair
The correct factor pair is $-1$ and $-7$, since:
- $-1 \times -7 = 7$
- $-1 + (-7) = -8$

#### Step 6: Write the factored form
Now that we have the correct factor pair, we can write the quadratic expression in its factored form:
$$
x^2 - 8x + 7 = (x - 1)(x - 7)
$$

#### Step 7: Solve for $x$
To find the $x$-intercepts, set each factor to zero:
$$
x - 1 = 0 \quad \text{or} \quad x - 7 = 0
$$
Thus, the solutions are:
$$
x = 1 \quad \text{or} \quad x = 7
$$

#### Final Answer
The $x$-intercepts of $f(x) = -x^2 + 8x - 7$ are the points:
$$
(1,0) \quad \text{and} \quad (7,0)
$$


## JIT 22. Radical Expressions.

$$\sqrt{ab}=\sqrt{a}\sqrt{b}$$

$$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$$

- Some square roots have whole number answers, when the numbers involve perfect squares:

$\sqrt{36}=6$

$\sqrt{100}=10$

$\sqrt{81}=9$

$\sqrt{\dfrac{9}{25}}=\dfrac{\sqrt{9}}{\sqrt{25}}=\dfrac{3}{5}$

$\sqrt{\dfrac{49}{36}}=\dfrac{\sqrt{49}}{\sqrt{36}}=\dfrac{7}{6}$

- Sometimes only certain factors of the square root involve perfect squares:

Examples.

$\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}$

$\sqrt{18}=\sqrt{9}\sqrt{2}=3\sqrt{2}$

$\sqrt{48}=\sqrt{16}\sqrt{3}=4\sqrt{3}$

$\sqrt{54}=\sqrt{9}\sqrt{6}=3\sqrt{6}$

$\sqrt{40}=\sqrt{4}\sqrt{10}=2\sqrt{10}$

$\sqrt{60}=\sqrt{4}\sqrt{15}=2\sqrt{15}$

$\sqrt{98}=\sqrt{49}\sqrt{2}=7\sqrt{2}$

## Quadratic Formula
- Sometimes quadratics are too difficult to factor.

- Then $ax^2 + bx + c = 0$ can be solved with the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

---

### Example 13. $x^2 + x - 7 = 0$

- No nice and easy way to factor.

$$
a = 1, \, b = 1, \, c = -7
$$

\begin{align*}
x &= \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-7)}}{2(1)} \\
x &= \frac{-1 \pm \sqrt{1 + 28}}{2} \\
x &= \frac{-1 \pm \sqrt{29}}{2}
\end{align*}

The two solutions are $x=-\dfrac{1}{2} + \dfrac{\sqrt{29}}{2}$ and $x=-\dfrac{1}{2} - \dfrac{\sqrt{29}}{2}$

---

### Example 14. Given the equation: $$x^2 + 3x + 1 = 0$$

Using the quadratic formula:

$$
a = 1, \, b = 3, \, c = 1
$$

\begin{align*}
x &= \frac{-3 \pm \sqrt{(3)^2 - 4(1)(1)}}{2(1)} \\
x &= \frac{-3 \pm \sqrt{9 - 4}}{2} \\
x &= \frac{-3 \pm \sqrt{5}}{2}
\end{align*}

The two solutions are $x=-\dfrac{3}{2} + \dfrac{\sqrt{5}}{2}$ and $x=-\dfrac{3}{2} - \dfrac{\sqrt{5}}{2}$

---

- Sometimes we have no REAL solutions - if we have a square root of a negative.

- We write $i=\sqrt{-1}$.

---

### Example 15 $x^2-x+5=0$

$$a=1, b=-1, c=5$$

$$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}$$

$$x=\frac{1\pm\sqrt{1-20}}{2}$$

$$x=\frac{1\pm\sqrt{-19}}{2}$$

$$x=\frac{1\pm\sqrt{19}\sqrt{-1}}{2}$$

$$x=\frac{1\pm\sqrt{19}i}{2}$$

The two solutions are $x=\dfrac{1}{2} + \dfrac{\sqrt{19}}{2}i$ and $x=\dfrac{1}{2} - \dfrac{\sqrt{19}}{2}i$.

- We still say there are no REAL solutions.

- In this case the graph of $y=ax^2+bx+c$ has no (real) $x$ intercepts.

---

- Sometimes we have some simplification to do:

---

### Example 16 $2x^2+2x-5=0$

$$a=2, b=2, c=-5$$

$$x=\frac{-2\pm\sqrt{2^2-4(2)(-5)}}{2(2)}$$

$$x=\frac{-2\pm\sqrt{4+40}}{4}$$

$$x=\frac{-2\pm\sqrt{44}}{4}$$

Here, $\sqrt{44}$ can be simplified as $\sqrt{(4)(11)}=\sqrt{4}\sqrt{11}=2\sqrt{11}$:

$$x=\frac{-2\pm 2\sqrt{11}}{4}$$
$$x=\frac{-2}{4} \pm \dfrac{2\sqrt{11}}{4}$$
$$x=-\frac{1}{2} \pm \dfrac{\sqrt{11}}{2}$$

The two solutions are $x=-\dfrac{1}{2} + \dfrac{\sqrt{11}}{2}i$ and $x=-\dfrac{1}{2} - \dfrac{\sqrt{11}}{2}i$.

---

### Example 17. $3x^2+2x-3=0$

#### Step 1: Identify the coefficients
For the quadratic equation $3x^2 + 2x - 3 = 0$, we have:
$$
a = 3, \, b = 2, \, c = -3
$$

#### Step 2: Use the quadratic formula
The quadratic formula is:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
Substitute the values of $a$, $b$, and $c$:
$$
x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-3)}}{2(3)}
$$

#### Step 3: Simplify under the square root
$$
x = \frac{-2 \pm \sqrt{4 + 36}}{6}
$$
$$
x = \frac{-2 \pm \sqrt{40}}{6}
$$

#### Step 4: Simplify the square root
We can simplify $\sqrt{40}$ as:
$$
\sqrt{40} = \sqrt{(4)(10)} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}
$$
Thus, we have:
$$
x = \frac{-2 \pm 2\sqrt{10}}{6}=\dfrac{-2}{6} \pm \dfrac{2\sqrt{10}}{6}=-\dfrac{1}{3} \pm \dfrac{\sqrt{10}}{3}
$$

#### Final Answer:
The solutions are:
$$
x = -\dfrac{1}{3} + \dfrac{\sqrt{10}}{3} \quad \text{or} \quad -\dfrac{1}{3} - \dfrac{\sqrt{10}}{3}
$$

---

### Example 18. $5x^2+2x=1$

#### Step 1: Rearrange the equation
Move the $1$ to the left-hand side to set the equation equal to $0$:
$$
5x^2 + 2x - 1 = 0
$$

#### Step 2: Identify the coefficients
For the quadratic equation $5x^2 + 2x - 1 = 0$, we have:
$$
a = 5, \, b = 2, \, c = -1
$$

#### Step 3: Use the quadratic formula
The quadratic formula is:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
Substitute the values of $a$, $b$, and $c$:
$$
x = \frac{-2 \pm \sqrt{2^2 - 4(5)(-1)}}{2(5)}
$$

#### Step 4: Simplify under the square root
$$
x = \frac{-2 \pm \sqrt{4 + 20}}{10}
$$
$$
x = \frac{-2 \pm \sqrt{24}}{10}
$$

#### Step 5: Simplify the square root
We can simplify $\sqrt{24}$ as:
$$
\sqrt{24} = \sqrt{(4)(6)} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}
$$
Thus, we have:
$$
x = \frac{-2 \pm 2\sqrt{6}}{10}=\dfrac{-2}{10} \pm \dfrac{2\sqrt{6}}{10}=-\dfrac{1}{5} \pm \dfrac{\sqrt{6}}{5}
$$

### Final Answer:
The solutions are:
$$
x = -\dfrac{1}{5} + \dfrac{\sqrt{6}}{5} \quad \text{or} \quad x = -\dfrac{1}{5} - \dfrac{\sqrt{6}}{5}
$$