## Precalculus: Mathematics for Calculus, 7th Edition

$x=-1\pm\dfrac{2}{3}\sqrt{6}$
$3x^{2}+6x-5=0$ Use the quadratic formula to solve this equation. This formula is: $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ For this particular equation, $a=3$, $b=6$ and $c=-5$ $x=\dfrac{-6\pm\sqrt{(6)^{2}-4(3)(-5)}}{2(3)}=\dfrac{-6\pm\sqrt{36+60}}{6}=\dfrac{-6\pm\sqrt{96}}{6}$ $x=\dfrac{-6\pm4\sqrt{6}}{6}=-1\pm\dfrac{2}{3}\sqrt{6}$