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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+9y^2-6y+96y-64)-4=0

We get rid of parentheses

9y^2-6y+96y-64-4=0

We add all the numbers together, and all the variables

9y^2+90y-68=0

a = 9; b = 90; c = -68;
Δ = b2-4ac
Δ = 902-4·9·(-68)
Δ = 10548
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:

$\sqrt{\Delta}=\sqrt{10548}=\sqrt{36*293}=\sqrt{36}*\sqrt{293}=6\sqrt{293}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-6\sqrt{293}}{2*9}=\frac{-90-6\sqrt{293}}{18}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+6\sqrt{293}}{2*9}=\frac{-90+6\sqrt{293}}{18}
$