X^2+40+4x=180

Solution for X^2+40+4x=180 equation:

X^2+40+4X=180

We move all terms to the left:

X^2+40+4X-(180)=0

We add all the numbers together, and all the variables

X^2+4X-140=0

a = 1; b = 4; c = -140;
Δ = b2-4ac
Δ = 42-4·1·(-140)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-24}{2*1}=\frac{-28}{2}
=-14
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+24}{2*1}=\frac{20}{2}
=10
$