Answer
$x=3$ or $x=-\displaystyle \frac{3}{2}$
Work Step by Step
We solve:
$\frac{2}{3}x^2-x-3=0$
We multiply by $3$:
$3(\displaystyle \frac{2}{3}x^{2}-x-3)=3*0$
$2x^{2}-3x-9=0$
We solve using the quadratic formula ($a=2,\ b=-3,\ c=-9$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\displaystyle \frac{-(-3)\pm\sqrt{(-3)^{2}-4*2*-9}}{2*2}$
$x=\displaystyle \frac{3\pm\sqrt{9+72}}{4}$
$x=\frac{3\pm\sqrt{81}}{4}$
$x=\frac{3\pm 9}{4}$
$x=\displaystyle \frac{3+9}{4}$ or $x=\displaystyle \frac{3-9}{4}$
$x=3$ or $x=-\displaystyle \frac{3}{2}$
