# How do you determine the solutions to 3x^2+10x-5=0?

Jan 12, 2017

$\frac{- 10 \pm 2 \sqrt{10}}{3}$
$y = 3 {x}^{2} + 10 x - 5 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 100 + 60 = 160$ --> $d = \pm 4 \sqrt{10}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{20}{6} \pm \frac{4 \sqrt{10}}{6} = \frac{- 10 \pm 2 \sqrt{10}}{3}$