Solving $\frac{x^2 + k^2 - 2xk}{x^2 + c^2 - 2xc} = e^{-2yg}\frac{k^2}{c^2}$ $$\frac{x^2 + k^2 - 2xk}{x^2 + c^2 - 2xc} = e^{-2yg}\frac{k^2}{c^2}$$
I am attempting to solve for $x$. I've been at it for a while and have reached to above equation, but now it seems to get quite messy. Any help?
Important:
$k = \frac{a+g}{2E}$, 
$c = \frac{a-g}{2E}$,
$g = \sqrt{a^2 + 4E}$.
What has yet to be defined are just real constants (i.e, $E$, $a$, $y$)

It helps to note that everything is pretty much a constant, hence we may reduce as follows:
$$\frac{x^2-2kx+k^2}{x^2-2cx+c^2}=P$$
where $P=e^{-2yg}\frac{k^2}{c^2}$.  Then notice some perfect squares:
$$x^2-2kx+k^2=(x-k)^2\\x^2-2cx+c^2=(x-c)^2$$
Thus, the problem further reduces:
$$\frac{x-k}{x-c}=\pm\sqrt P$$
Multiply both sides by $x-c$ to get
$$x-k=\pm\sqrt P(x-c)$$
$$\mp c\sqrt P-k=(\pm\sqrt P-1)x$$

$$x=\frac{\mp c\sqrt P-k}{\pm\sqrt P-1}$$

(watch the signs, as $\pm$ is always the opposite sign of $\mp$)
