3y^2+12y-12=0

Solution for 3y^2+12y-12=0 equation:

3y^2+12y-12=0

a = 3; b = 12; c = -12;
Δ = b2-4ac
Δ = 122-4·3·(-12)
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{2}}{2*3}=\frac{-12-12\sqrt{2}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{2}}{2*3}=\frac{-12+12\sqrt{2}}{6}
$