## College Algebra 7th Edition

$\displaystyle \frac{-2\pm\sqrt{7}}{3}$
$3x^{2}+4x-1=0$ We solve using the quadratic equation ($a=3, b=4, c=-1$): $\displaystyle x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-4\pm\sqrt{4^{2}-4*3*-1}}{2*-3}=\frac{-4\pm\sqrt{16+12}}{-6}=\frac{-4\pm\sqrt{28}}{-6}=\frac{-4\pm 2\sqrt{7}}{6}=\frac{-2\pm\sqrt{7}}{3}$