Solving $x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$. I have problem with this equation:
$$x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$$
Any ideas on beatiful solving?

$$
\begin{array}{l l l}
x-\sqrt{(x^2-36)}& =& {(x-6)^2\over 2x+12}\\
(2x+12)\left(x-\sqrt{(x^2-36)}\right)&=&(x-6)^2\\
12x+2x^2+(-12-2x)\sqrt{(x^2-36)}&=&(x-6)^2\\
(-12-2x)\sqrt{(x^2-36)}&=&(x-6)^2-12x-2x^2\\
(-12-2x)^2(x^2-36)&=&\left((x-6)^2-12x-2x^2\right)^2\\
4x^4+48x^3-1728x-5184&=&x^4+48x^3+504x^2-1728x+1296\\
3x^4-504x^2-6480&=&0\\
\text{Substitute }y=x^2\\
3y^2-504y-6480&=&0\\
3(y-180)(y+12)&=&0\\
(y-180)(y+12)&=&0\\
\end{array}
$$
We now split into 2 equations
$$y-180=0\text{ or }y+12=0$$
$$y=180\text{ or }y=-12$$
$$x^2=180\text{ or }x^2=-12$$
$\sqrt{-12}$ has no solutions
$$x=\pm\sqrt{180}=\pm6\sqrt{5}$$
