3y(y+4)=24

Solution for 3y(y+4)=24 equation:

3y(y+4)=24

We move all terms to the left:

3y(y+4)-(24)=0

We multiply parentheses

3y^2+12y-24=0

a = 3; b = 12; c = -24;
Δ = b2-4ac
Δ = 122-4·3·(-24)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{3}}{2*3}=\frac{-12-12\sqrt{3}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{3}}{2*3}=\frac{-12+12\sqrt{3}}{6}
$