x^2+3x+40+x^2+11x+20=180

Solution for x^2+3x+40+x^2+11x+20=180 equation:

x^2+3x+40+x^2+11x+20=180

We move all terms to the left:

x^2+3x+40+x^2+11x+20-(180)=0

We add all the numbers together, and all the variables

2x^2+14x-120=0

a = 2; b = 14; c = -120;
Δ = b2-4ac
Δ = 142-4·2·(-120)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-34}{2*2}=\frac{-48}{4}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+34}{2*2}=\frac{20}{4}
=5
$