# How do you solve 5y^2-180=0?

$5 {y}^{2} - 180 = 0 \implies 5 \cdot \left({y}^{2} - 36\right) = 0 \implies 5 \cdot \left(y - 6\right) \left(y + 6\right) = 0$
${y}_{1} = 6$ and ${y}_{2} = - 6$