Answer
The solutions are $x=-\dfrac{3}{16}\pm\dfrac{3\sqrt{129}}{16}$
Work Step by Step
$\dfrac{2}{3}x^{2}+\dfrac{1}{4}x=3$
Multiply the whole equation by $12$:
$12\Big(\dfrac{2}{3}x^{2}+\dfrac{1}{4}x=3\Big)$
$8x^{2}+3x=36$
Take $36$ to the left side:
$8x^{2}+3x-36=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=8$, $b=3$ and $c=-36$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-3\pm\sqrt{3^{2}-4(8)(-36)}}{2(8)}=\dfrac{-3\pm\sqrt{9+1152}}{16}=...$
$...=\dfrac{-3\pm\sqrt{1161}}{16}=\dfrac{-3\pm3\sqrt{129}}{16}=-\dfrac{3}{16}\pm\dfrac{3\sqrt{129}}{16}$
The solutions are $x=-\dfrac{3}{16}\pm\dfrac{3\sqrt{129}}{16}$
