Question

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

Answer 1

`y_1=-6`, `y_2=0` and `y_3=5/2`

Explanation:

`2y^3+7y^2-30y=0`

`y*(2y^2+7y-30)=0`

`y*(2y^2+12y-5y-30)=0`

`y*(y+6)*(2y-5)=0`

Thus, `y_1=-6`, `y_2=0` and `y_3=5/2`