Answer
$-\frac{3\pm\sqrt{13}}{4}$
Work Step by Step
First we multiply through by 2:
$2x^{2}+3x-\frac{1}{2}=0$
Then we solve using the quadratic formula:
$4x^{2}+6x-1=0$
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-6\pm\sqrt{6^{2}-4(4)(-1)}}{2(4)}=\frac{-6\pm\sqrt{52}}{8}=-\frac{3\pm\sqrt{13}}{4}$