Answer
$\displaystyle \frac{x^{2}+40x-25}{(x+5)(x-4)},\qquad x\neq-5,4$
Work Step by Step
Factor each denominator.
$x^{2}+x-20= (x+5)(x-4)$
... two factors of $-20$ with sum $+1$ are $+5$ and $-4.$
$LCD=(x+5)(x-4)$
$...= \displaystyle \frac{6x^{2}+17x-40}{(x+5)(x-4)}+\frac{3}{x-4}-\frac{5x}{x+5}$
$= \displaystyle \frac{6x^{2}+17x-40}{(x+5)(x-4)}+\frac{3}{x-4}\times\frac{x+5}{x+5}-\frac{5x}{x+5}\times\frac{x-4}{x-4}$
$=\displaystyle \frac{6x^{2}+17x-40}{(x+5)(x-4)} +\frac{3x+15}{(x+5)(x-4)}-\frac{5x^{2}-4x}{(x+5)(x-4)}$
$=\displaystyle \frac{6x^{2}+17x-40+3x+15-5x^{2}+20x}{(x+5)(x-4)}$
= $\displaystyle \frac{x^{2}+40x-25}{(x+5)(x-4)},\qquad x\neq-5,4$
