180=(y-18)(y+12)

Solution for 180=(y-18)(y+12) equation:

180=(y-18)(y+12)

We move all terms to the left:

180-((y-18)(y+12))=0

We multiply parentheses ..

-((+y^2+12y-18y-216))+180=0


We calculate terms in parentheses: -((+y^2+12y-18y-216)), so:

(+y^2+12y-18y-216)

We get rid of parentheses

y^2+12y-18y-216

We add all the numbers together, and all the variables

y^2-6y-216

Back to the equation:

-(y^2-6y-216)


We get rid of parentheses

-y^2+6y+216+180=0

We add all the numbers together, and all the variables

-1y^2+6y+396=0

a = -1; b = 6; c = +396;
Δ = b2-4ac
Δ = 62-4·(-1)·396
Δ = 1620
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:

$\sqrt{\Delta}=\sqrt{1620}=\sqrt{324*5}=\sqrt{324}*\sqrt{5}=18\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18\sqrt{5}}{2*-1}=\frac{-6-18\sqrt{5}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18\sqrt{5}}{2*-1}=\frac{-6+18\sqrt{5}}{-2}
$