Why $e=g \pm \sqrt{g}$ when $ \left( g-e \right)^2=g$?

Question

I have a problem that involves finding $e$ such that $\left( g-e \right)^2=g$. Maxima tells me that $e=g \pm \sqrt{g}$, but I can't work on that equation to get this result. Actually, I can't go past $2ge - e^2 = g^2 - g$. Can someone show me the algebraic steps to get that result?

Answer 1

We have $(g-e)^2 = g$. Taking square roots, $(g-e) = \pm \sqrt{g}$. Rearranging terms gives $ e = g \pm \sqrt{g}$.

Don't expand the left side; this makes it more difficult.

Answer 2

If $(g-e)^2=g$, then either $g-e=\sqrt g$, or $g-e=-\sqrt g$. Now solve these equations for $e$ to get the two possibilities.