# How do you solve 2y^3+7y^2-30y=0?

Mar 24, 2018

${y}_{1} = - 6$, ${y}_{2} = 0$ and ${y}_{3} = \frac{5}{2}$

#### Explanation:

$2 {y}^{3} + 7 {y}^{2} - 30 y = 0$

$y \cdot \left(2 {y}^{2} + 7 y - 30\right) = 0$

$y \cdot \left(2 {y}^{2} + 12 y - 5 y - 30\right) = 0$

$y \cdot \left(y + 6\right) \cdot \left(2 y - 5\right) = 0$

Thus, ${y}_{1} = - 6$, ${y}_{2} = 0$ and ${y}_{3} = \frac{5}{2}$