#### Answer

{$\dfrac{-3}{2} ,-1$}

#### Work Step by Step

Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
As we are given that $2x^2+8x=3x-3$
or, $2x^2+8x-3x+3=0$
This implies $2x^2+5x+3=0$
Thus, $x=\dfrac{-(5) \pm \sqrt{(5)^2-4(2)(3)}}{2(2)}$
or, $x=\dfrac{-5 \pm \sqrt{1}}{4} \implies x= \dfrac{-5 \pm 1}{4}$
or, $x=\dfrac{-5 - 1}{4},\dfrac{-5 + 1}{4}$
Our solution set is: {$\dfrac{-3}{2} ,-1$}