X^2+20x-40=180

Solution for X^2+20x-40=180 equation:

X^2+20X-40=180

We move all terms to the left:

X^2+20X-40-(180)=0

We add all the numbers together, and all the variables

X^2+20X-220=0

a = 1; b = 20; c = -220;
Δ = b2-4ac
Δ = 202-4·1·(-220)
Δ = 1280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{5}}{2*1}=\frac{-20-16\sqrt{5}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{5}}{2*1}=\frac{-20+16\sqrt{5}}{2}
$