How to solve this for q?

Question

So far in this, I understand how to get $$(gamma^2/4b^2)(q1)$$, but not the bit before the '+' on the bottom line. Can anyone break it down for me? Thanks!

Answer 1

from your last equation we get $$q_1\left(\frac{4b^2-\gamma^2}{4b^2}\right)=\frac{1}{4b^2}\left(2ab-2bc-a\gamma+c\gamma\right)$$ and cancelling $4b^2$ and dividing by $$4b^2-\gamma^2\neq 0$$ we get $$q_1=\frac{2ab-2bc-a\gamma+c\gamma}{4b^2-\gamma^2}$$