0=180-x^2-4x+12

Solution for 0=180-x^2-4x+12 equation:

0=180-x^2-4x+12

We move all terms to the left:

0-(180-x^2-4x+12)=0

We add all the numbers together, and all the variables

-(180-x^2-4x+12)=0

We get rid of parentheses

x^2+4x-180-12=0

We add all the numbers together, and all the variables

x^2+4x-192=0

a = 1; b = 4; c = -192;
Δ = b2-4ac
Δ = 42-4·1·(-192)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-28}{2*1}=\frac{-32}{2}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+28}{2*1}=\frac{24}{2}
=12
$