x^2+40+3x+10=180

Solution for x^2+40+3x+10=180 equation:

x^2+40+3x+10=180

We move all terms to the left:

x^2+40+3x+10-(180)=0

We add all the numbers together, and all the variables

x^2+3x-130=0

a = 1; b = 3; c = -130;
Δ = b2-4ac
Δ = 32-4·1·(-130)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-23}{2*1}=\frac{-26}{2}
=-13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+23}{2*1}=\frac{20}{2}
=10
$