Answer
$x=-1\displaystyle \pm\frac{2\sqrt{6}}{3}$
Work Step by Step
We solve by completing the square:
$3x^{2}+6x-5=0$
$x^{2}+2x-\frac{5}{3}=0$
$x^{2}+2x=\frac{5}{3}$
$x^{2}+2x+1=\frac{5}{3}+1$
$(x+1)^{2}=\frac{8}{3}$
$x+1=\pm\sqrt{\frac{8}{3}}=\pm\sqrt{\frac{4*2*3}{3*3}}=\pm\frac{2\sqrt{6}}{3}$
$x=-1\displaystyle \pm\frac{2\sqrt{6}}{3}$
