Question: I'm trying to write a programme that draws an IV curve for a diode, using an expanded version of the Shockley diode equation that uses extra variables, but now the 'I' term appears on both sides of the equation and I can't work out how to rearrange it to do so. I understand that these things can be solved iteratively, but would prefer the rearrangement to make I the subject and only appear on one side.

$$I=I\_0e^{\frac{q(V-IR\_{se})}{nkT}}+\frac{V-IR\_{se}}{R\_{sh}} -I\_L$$

Answer: 
\begin{array}{r}
 (x+2)^4 = && \star x^4 &+ \star x^3 &+ 24x^2 &+ 32x &+ 16 \\
 (x+3)^5 = &\star x^5 &+ \star x^4 &+ \star x^3 &+ 270x^2 &+ 405x &+ 243
\end{array}

\begin{array}{c|c}
 \times &16 & 32x &24x^2 &\cdots \\
\hline
 243 & \star & \star x & 5832x^2 &\cdots \\
 405x & \star x & 12960x^2 & \star x^3 &\cdots \\
 270x^2 & 4320 x^2 & \star x^3 & \star x^4 &\cdots \\
 \vdots & \vdots & \vdots& \vdots \\
\hline
\end{array}
Wouldn't the $x$ term of one and the $x$ term of the other combine in the product to produce an $x^2$ term in the result?