College Algebra (10th Edition)

$x=1$ or $x=-\displaystyle \frac{2}{3}$
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 $x=-\displaystyle \frac{2}{3}$