180=-x^2+12x+160

Solution for 180=-x^2+12x+160 equation:

180=-x^2+12x+160

We move all terms to the left:

180-(-x^2+12x+160)=0

We get rid of parentheses

x^2-12x-160+180=0

We add all the numbers together, and all the variables

x^2-12x+20=0

a = 1; b = -12; c = +20;
Δ = b2-4ac
Δ = -122-4·1·20
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*1}=\frac{20}{2}
=10
$