Answer
$\dfrac{2(x+3)}{5(x-5)}$
Work Step by Step
Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, then the given expression, $
\dfrac{2x^2-4x-30}{5x^2-40x-75}\div\dfrac{x^2-8x+15}{x^2-6x+9}
$, simplifies to
\begin{array}{l}\require{cancel}
\dfrac{2x^2-4x-30}{5x^2-40x-75}\cdot\dfrac{x^2-6x+9}{x^2-8x+15}
\\\\=
\dfrac{2(x^2-2x-15)}{5(x^2-8x-15)}\cdot\dfrac{(x-3)(x-3)}{(x-5)(x-3)}
\\\\=
\dfrac{2(x-5)(x+3)}{5(x-5)(x-3)}\cdot\dfrac{(x-3)(x-3)}{(x-5)(x-3)}
\\\\=
\dfrac{2(\cancel{x-5})(x+3)}{5(\cancel{x-5})(\cancel{x-3})}\cdot\dfrac{(\cancel{x-3})(\cancel{x-3})}{(x-5)(\cancel{x-3})}
\\\\=
\dfrac{2(x+3)}{5(x-5)}
.\end{array}
