Algebra help

hint: apply radical rule!

I'll post a solution later, but my answer is: $\frac{\sqrt{2h-2x}}{h}-i \frac{\sqrt2\sqrt x}{h}$

(sqrt(-2(h+x))-sqrt(-2x))/h

$\frac{{\sqrt{-2(h+x)}-\sqrt{-2x}} }{h}    \\ \frac{{\sqrt{-1}\sqrt{2(h+x)}-\sqrt{-1}\sqrt{2x}} }{h}    \\ \frac{\sqrt{-1}\left[\sqrt{2(h+x)}-\sqrt{2x}\right]}{h}    \\ \frac{\left[\sqrt{2h+2x}-\sqrt{2x}\;\right]\;i}{h}    \\ or\\ \frac{\left[\sqrt{h+x}-\sqrt{x}\;\right]\sqrt2\;i}{h}    \\ $

the final answer really depends on what you are doing it for.