x^2+3x+x+120=180

Solution for x^2+3x+x+120=180 equation:

x^2+3x+x+120=180

We move all terms to the left:

x^2+3x+x+120-(180)=0

We add all the numbers together, and all the variables

x^2+4x-60=0

a = 1; b = 4; c = -60;
Δ = b2-4ac
Δ = 42-4·1·(-60)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*1}=\frac{-20}{2}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*1}=\frac{12}{2}
=6
$