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

{$-1,\dfrac{-3}{5}$}
Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ Given: $5x^2+8x=-3$ It can be written as: $5x^2+8x+3=0$ Thus, $x=\dfrac{-(8) \pm \sqrt{(8)^2-4(5)(3)}}{2(5)}$ or, $x=\dfrac{-8 \pm 2}{10}$ or, $x=\dfrac{-8 + 2}{10}=\dfrac{-3}{5}$ and $x=\dfrac{-8-2}{10}=-1$ Our solution set is: {$-1,\dfrac{-3}{5}$}