# How do you solve 3y^2+2y-1=0?

Mar 30, 2018

$y = - 1 , \frac{1}{3}$
$3 {y}^{2} + 3 y - y - 1 = 0$ find two integers that ADD to 2, MULTIPLY to -3 $\left(3 \cdot - 1\right)$
$3 y \left(y + 1\right) - \left(y + 1\right)$ factor out common factor
$\left(3 y - 1\right) \left(y + 1\right)$
$3 y - 1 = 0$ OR $y + 1 = 0$
$y = \frac{1}{3}$ OR $y = - 1$