Answer
$x=\left\{ \dfrac{-3-3\sqrt{129}}{16},\dfrac{-3+3\sqrt{129}}{16} \right\}
$
Work Step by Step
Using the properties of equality, the given equation, $
\dfrac{2}{3}x^2+\dfrac{1}{4}x=3
,$ is equivalent to
\begin{array}{l}\require{cancel}
12\left(\dfrac{2}{3}x^2+\dfrac{1}{4}x\right)=(3)12
\\\\
8x^2+3x=36
\\\\
8x^2+3x-36=0
.\end{array}
The equation above has $a=8$, $b=3$, and $c=-36$.
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions to the equation above are
\begin{array}{l}\require{cancel}
x=\dfrac{-3\pm\sqrt{3^2-4(8)(-36)}}{2(8)}
\\\\
x=\dfrac{-3\pm\sqrt{9+1152}}{16}
\\\\
x=\dfrac{-3\pm\sqrt{1161}}{16}
\\\\
x=\dfrac{-3\pm\sqrt{9\cdot129}}
\\\\
x=\dfrac{-3\pm\sqrt{(3)^2\cdot129}}{16}
\\\\
x=\dfrac{-3\pm3\sqrt{129}}{16}
.\end{array}
Hence, the solutions are $
x=\left\{ \dfrac{-3-3\sqrt{129}}{16},\dfrac{-3+3\sqrt{129}}{16} \right\}
.$
