How to solve this equation for c? I want to solve the equation
$$
-\frac{a}{2}\left(c+\sqrt{c^2+4}\right)=-\frac{a-1}{2}\left(c-\sqrt{c^2+4}\right)
$$
for $c$, where $a$ is just a constant.
What I get is
$$
\frac{c-\sqrt{c^2+4}}{c+\sqrt{c^2+4}}=\frac{a}{a-1}.
$$
I think there now is some "trick" to solve this for $c$.

By distributing, 
$$-\frac{ac}{2}-\frac{a\sqrt{c^2+4}}{2}=-\frac{ac}{2}+\frac{a\sqrt{c^2+4}}{2}+\frac{c}{2}-\frac{\sqrt{c^2+4}}{2}$$
Well,
$$0=(a-\frac{1}{2})\sqrt{c^2+4}+\frac{c}{2}$$
$$\frac{1}{1-2a}=\frac{\sqrt{c^2+4}}{c}$$ with assumption that $c \ne0.$
Squaring, $$\frac{1}{(1-2a)^2}=\frac{c^2+4}{c^2}=1+\frac{4}{c^2}$$
$$\frac{4}{c^2}=\frac{4a-4a^2}{(1-2a)^2}$$
By fliping, and multiplying both sides by $4$,
$$c^2=\frac{(1-2a)^2}{a-a^2}$$
You now have $c$ without rationalising surds.
