(2y+32)*(3y-2)=4

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+6y^2-4y+96y-64)-4=0

We get rid of parentheses

6y^2-4y+96y-64-4=0

We add all the numbers together, and all the variables

6y^2+92y-68=0

a = 6; b = 92; c = -68;
Δ = b2-4ac
Δ = 922-4·6·(-68)
Δ = 10096
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:

$\sqrt{\Delta}=\sqrt{10096}=\sqrt{16*631}=\sqrt{16}*\sqrt{631}=4\sqrt{631}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-4\sqrt{631}}{2*6}=\frac{-92-4\sqrt{631}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+4\sqrt{631}}{2*6}=\frac{-92+4\sqrt{631}}{12}
$