(3y+1)(2y+2)48=180

Solution for (3y+1)(2y+2)48=180 equation:

(3y+1)(2y+2)48=180

We move all terms to the left:

(3y+1)(2y+2)48-(180)=0

We multiply parentheses ..

(+6y^2+6y+2y+2)48-180=0

We multiply parentheses

288y^2+288y+96y+96-180=0

We add all the numbers together, and all the variables

288y^2+384y-84=0

a = 288; b = 384; c = -84;
Δ = b2-4ac
Δ = 3842-4·288·(-84)
Δ = 244224
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:

$\sqrt{\Delta}=\sqrt{244224}=\sqrt{2304*106}=\sqrt{2304}*\sqrt{106}=48\sqrt{106}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(384)-48\sqrt{106}}{2*288}=\frac{-384-48\sqrt{106}}{576}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(384)+48\sqrt{106}}{2*288}=\frac{-384+48\sqrt{106}}{576}
$