# How do you solve 2y^2 + 34y +132 = 0?

Apr 9, 2018

$2 {y}^{2} + 34 y + 132 = 0$

$\implies {y}^{2} + 17 y + 66 = 0$

$\implies {y}^{2} + 6 y + 11 y + 66 = 0$

$\implies y \left(y + 6\right) + 11 \left(y + 6\right) = 0$

$\implies \left(y + 11\right) \left(y + 6\right) = 0$

Therefore either $y + 11 = 0$ or $y + 6 = 0$
thats is $y = - 11$ or $y = - 6$