Answer
{$\dfrac{-3 - \sqrt{41}}{8},\dfrac{-3 + \sqrt{41}}{8}$}
Work Step by Step
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: {$\dfrac{-3 - \sqrt{41}}{8},\dfrac{-3 + \sqrt{41}}{8}$}
