# How do you solve the equation 3x^2+6x-1=8?

You want an equation like $a {x}^{2} + b x + c = 0$, so we subtract 8 from both sides. We now get $3 {x}^{2} + 6 x - 9 = 0$, so we find $a = 3 , b = 6$ and $c = - 9$.
$D = {b}^{2} - 4 a c = {6}^{2} - 4 \cdot 3 \cdot - 9 = 36 + 108 = 144.$
We know that $x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{- 6 \pm \sqrt{144}}{2 \cdot 3} = \frac{- 6 \pm 12}{6} = - 1 \pm 2$, so $x = 1 \vee x = - 3$.