Question

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

Answer 1

`y=-1, 1/3`

Explanation:

`3y^2+3y-y-1=0` find two integers that ADD to 2, MULTIPLY to -3 `(3*-1)`
`3y(y+1)-(y+1)` factor out common factor
`(3y-1)(y+1)`
`3y-1 = 0` OR `y+1=0`
`y=1/3` OR `y=-1`