## Intermediate Algebra for College Students (7th Edition)

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