## College Algebra 7th Edition

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