4y(y-12)=36

Solution for 4y(y-12)=36 equation:

4y(y-12)=36

We move all terms to the left:

4y(y-12)-(36)=0

We multiply parentheses

4y^2-48y-36=0

a = 4; b = -48; c = -36;
Δ = b2-4ac
Δ = -482-4·4·(-36)
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-24\sqrt{5}}{2*4}=\frac{48-24\sqrt{5}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+24\sqrt{5}}{2*4}=\frac{48+24\sqrt{5}}{8}
$