3y^2+24y-36=0

Solution for 3y^2+24y-36=0 equation:

3y^2+24y-36=0

a = 3; b = 24; c = -36;
Δ = b2-4ac
Δ = 242-4·3·(-36)
Δ = 1008
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:

$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{7}}{2*3}=\frac{-24-12\sqrt{7}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{7}}{2*3}=\frac{-24+12\sqrt{7}}{6}
$