## College Algebra 7th Edition

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