100+20x=x^2

Solution for 100+20x=x^2 equation:

100+20x=x^2

We move all terms to the left:

100+20x-(x^2)=0

determiningTheFunctionDomain
-x^2+20x+100=0

We add all the numbers together, and all the variables

-1x^2+20x+100=0

a = -1; b = 20; c = +100;
Δ = b2-4ac
Δ = 202-4·(-1)·100
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{2}}{2*-1}=\frac{-20-20\sqrt{2}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{2}}{2*-1}=\frac{-20+20\sqrt{2}}{-2}
$