(180-x^2)+(4x+36)+(6x-12)=180

Solution for (180-x^2)+(4x+36)+(6x-12)=180 equation:

(180-x^2)+(4x+36)+(6x-12)=180

We move all terms to the left:

(180-x^2)+(4x+36)+(6x-12)-(180)=0

We get rid of parentheses

-x^2+4x+6x+180+36-12-180=0

We add all the numbers together, and all the variables

-1x^2+10x+24=0

a = -1; b = 10; c = +24;
Δ = b2-4ac
Δ = 102-4·(-1)·24
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*-1}=\frac{-24}{-2}
=+12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*-1}=\frac{4}{-2}
=-2
$