Answer
$x_{1}= \dfrac{4}{3}$ and $x_{2} = -2$
Work Step by Step
Given $3x^2+2x=8 \longrightarrow 3x^2+2x-8=0$
$a= 3, \ b=2, \ c=-8$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have:
$\dfrac{-2 \pm \sqrt{2^2-4 \times 3\times (-8)}}{2 \times 3} = \dfrac{-2 \pm \sqrt{4+96}}{6} = \dfrac{-2 \pm \sqrt{100}}{6} = \dfrac{-2 \pm 10}{6} = \dfrac{-1 \pm 5}{3}$
Therefore the solutions are $x_{1}= \dfrac{-1 + 5}{3} =\dfrac{4}{3}$ and $x_{2} = \dfrac{-1 - 5}{3} = \dfrac{-6}{3} = -2$
