Answer
$\dfrac{2x^{2}-8}{4x-8}=\dfrac{x+2}{2}$
Work Step by Step
$\dfrac{2x^{2}-8}{4x-8}$
Take out common factor $2$ in the numerator and common factor $4$ in the denominator.
$\dfrac{2x^{2}-8}{4x-8}=\dfrac{2(x^{2}-4)}{4(x-2)}=...$
Now, factor $x^{2}-4$ and simplify:
$...=\dfrac{2(x-2)(x+2)}{4(x-2)}=\dfrac{2(x+2)}{4}=\dfrac{x+2}{2}$