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: 
Using the binomial theorem we have:
$$(x+2)^4(x+3)^5 = \left( \sum\_{k=0}^{4}\binom{4}{k}x^k2^{4-k}\right) \left(\sum\_{j=0}^{5}\binom{5}{j}x^j3^{5-j} \right).$$

We get $x^2$ by taking an $x$ term from each sum or an $x^2$ and $x^0$ term from one or the other sum. So the coefficient of $x^2$ is $$\overbrace{\binom{4}{1}2^3\binom{5}{1}3^4}^{x\cdot x} + \overbrace{2^4\binom{5}{2}3^3}^{x^0\cdot x^2} + \overbrace{\binom{4}{2}2^23^5}^{x^2\cdot x^0}$$
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?