Answer
$=\frac{5(2x-5)}{x-5}, with$ $x\ne5$
Work Step by Step
Given : $\frac{5x^{2}+10x-15}{5-6x+x^{2}} \div \frac{2x^{2}+7x+3}{4x^{2}-8x-15}$
This becomes : $\frac{5x^{2}+10x-15}{5-6x+x^{2}} \times \frac{4x^{2}-8x-15}{2x^{2}+7x+3}=\frac{5(x-1)(x+3)}{(x-1)(x-5)} \times \frac{(2x+1)(2x-5)}{(x+3)(2x+1)}$
$=\frac{5(2x-5)}{x-5}$
(After dividing out the common factors $(x-1),(x+3)(2x+1)$)
