3y(2y-4)=32

Solution for 3y(2y-4)=32 equation:

3y(2y-4)=32

We move all terms to the left:

3y(2y-4)-(32)=0

We multiply parentheses

6y^2-12y-32=0

a = 6; b = -12; c = -32;
Δ = b2-4ac
Δ = -122-4·6·(-32)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{57}}{2*6}=\frac{12-4\sqrt{57}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{57}}{2*6}=\frac{12+4\sqrt{57}}{12}
$