Answer
$x=1$ or $x=-\displaystyle \frac{2}{3}$
Work Step by Step
We solve:
$\displaystyle \frac{3}{4}x^{2}-\frac{1}{4}x-\frac{1}{2}=0$
We multiply by $4$:
$4(\displaystyle \frac{3}{4}x^{2}-\frac{1}{4}x-\frac{1}{2})=4*0$
$3x^{2}-x-2=0$
We solve using the quadratic formula ($a=3,\ b=-1,\ c=-2$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\displaystyle \frac{-(-1)\pm\sqrt{(-1)^{2}-4*3*-2}}{2*3}$
$x=\displaystyle \frac{1\pm\sqrt{1+24}}{6}$
$x=\frac{1\pm\sqrt{25}}{6}$
$x=\frac{1\pm 5}{6}$
$x=\displaystyle \frac{1+5}{6}$ or $x=\displaystyle \frac{1-5}{6}$
$x=1$ or $x=-\displaystyle \frac{2}{3}$
