(x^2-2x)+(3x+108)=180

Solution for (x^2-2x)+(3x+108)=180 equation:

(x^2-2x)+(3x+108)=180

We move all terms to the left:

(x^2-2x)+(3x+108)-(180)=0

We get rid of parentheses

x^2-2x+3x+108-180=0

We add all the numbers together, and all the variables

x^2+x-72=0

a = 1; b = 1; c = -72;
Δ = b2-4ac
Δ = 12-4·1·(-72)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*1}=\frac{16}{2}
=8
$