The Steinheart-Heart equation:
$$
T = 1 / A+B ln(R) + C [ln(R)]^3
$$

Determing the 3 reference temperatures:
$$
1/T1=A+B ln(R_1) + C[ln(R_1)]3 \\
$$

$$
1/T2=A+B ln(R_2) + C[ln(R_2)]3
$$

$$
1/T3= A + B ln(R_3) + C[ln(R_3)]3
$$

Solving for coefficients:

$$
C=\frac{X_3 (Y_1-Y_2) - X_1(Y_1-Y_2) + Y_1 X_1 - Y_1 X_2 - Y_3 X_1 + Y_3X_2}
{({y_3})^{3} (y_1-Y_2) - (y_1)(y_2)^3 + (y_1)^4 - (y_1)^3(y_1−y_2) + (Y_3)(Y_2)^3 - (Y_3)(Y_1)^3}
$$

$$
B = \frac{(x_1-x_2) + C(y_2)^3-C(y_1)^3}{(Y_1-Y_2)}
$$

$$
A= x_2-By_2-C(y_2)^3
$$

---
Where:

<table class="fixed-align">
<tbody>
<tr>
<td>

$$
X_1 = 1/Tlow
$$

$$
X_2 = 1/Tmid
$$

$$
Y_1 = InRTIN
$$

$$
Y_2 = InRT
$$

$$
X_3 = 1/Thigh
$$

$$
Y_3 = In RThigh
$$

</td>

<td valign="top", valign="right">

$$
T_i = Kelvin Temperature = t_i (°C) + 273.15
$$

$$
Tlow = Low temperature calibration point
$$

$$
Tmid = Mid temperature calibration point
$$

$$
Thigh = High temperature calibration point
$$

$$
R1 = Resistance in ohms at temperature T_i
$$

</td>
</tr>
</tbody>
</table>