Question

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

Answer 1

`2y^2 + 34y +132 = 0`

`=> y^2 + 17y +66 = 0`

`=> y^2 + 6y +11y+66 = 0`

`=> y(y + 6) +11(y+6) = 0`

`=> (y+11)(y+6) = 0`

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