Answer
$-12, 2$
Work Step by Step
$\frac{x}{x-5} + \frac{5}{x+5} = \frac{-1}{x^{2}-25}$
Using $(a^{2}-b^{2}) = (a+b)(a-b)$ formula,
$x^{2}-25 = (x^{2}-5^{2}) = (x+5)(x-5)$
$\frac{x}{x-5} + \frac{5}{x+5} = \frac{-1}{(x+5)(x-5)}$
$x=5 $ or $x=-5$ makes the denominator zero. So,
$\frac{x}{x-5} + \frac{5}{x+5} = \frac{-1}{(x+5)(x-5)}$;$ x\ne -5, $ or $x \ne 5$
Multiply both sides by $(x+5)(x-5)$ to clear fractions.
$(x+5)(x-5)(\frac{x}{x-5} + \frac{5}{x+5}) =(x+5)(x-5)( \frac{-1}{(x+5)(x-5)})$
$(x+5)x+5(x-5) = -1 ; x\ne -5, $ or $x \ne 5$
$x^{2}+5x+5x-25=-1 $
$x^{2}+10x-25+1=0 $
$x^{2}+10x-24=0 $
By factoring,
$(x+12)(x-2)=0$
$x=-12$ or $x=2$
Solutions are $-12,2$