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

{$\dfrac{-3 - \sqrt{41}}{8},\dfrac{-3 + \sqrt{41}}{8}$}
Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ Given: $4x^2+3x=2$ This can be re-written as: $4x^2+3x-2=0$ Thus, $x=\dfrac{-(3) \pm \sqrt{(3)^2-4(4)(-2)}}{2(4)}$ or, $x=\dfrac{-3 \pm \sqrt{41}}{8}$ or, $x=\dfrac{-3 - \sqrt{41}}{8},\dfrac{-3 + \sqrt{41}}{8}$ Our solution set is: {$\dfrac{-3 - \sqrt{41}}{8},\dfrac{-3 + \sqrt{41}}{8}$}