x^2+20x=36

Solution for x^2+20x=36 equation:

x^2+20x=36

We move all terms to the left:

x^2+20x-(36)=0

a = 1; b = 20; c = -36;
Δ = b2-4ac
Δ = 202-4·1·(-36)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{34}}{2*1}=\frac{-20-4\sqrt{34}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{34}}{2*1}=\frac{-20+4\sqrt{34}}{2}
$