Answer
$\frac{5(2x-5)}{(x-5)}$
Work Step by Step
$\frac{5x^2+10x-15}{5-6x+x^2}\div\frac{2x^2+7x+3}{4x^2-8x-5}$
$\frac{5x^2+10x-15}{x^2-6x+5}*\frac{4x^2-8x-5}{2x^2+7x+3}$
$\frac{5x^2+10x-15}{x^2-6x+5}*\frac{4x^2-8x-5}{2x^2+7x+3}$
$\frac{5(x^2+2x-3)}{(x-1)(x-5)}*\frac{(2x+1)(2x-5)}{(2x+1)(x+3)}$
$\frac{5(x+3)(x-1)}{(x-1)(x-5)}*\frac{(2x+1)(2x-5)}{(2x+1)(x+3)}$
$\frac{5(x+3)}{(x-5)}*\frac{(2x-5)}{(x+3)}$
$\frac{5(x+3)(2x-5)}{(x-5)(x+3)}$
$\frac{5(2x-5)}{(x-5)}$
