x^2+20x+100=1

Solution for x^2+20x+100=1 equation:

x^2+20x+100=1

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2+20x+99=0

a = 1; b = 20; c = +99;
Δ = b2-4ac
Δ = 202-4·1·99
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2}{2*1}=\frac{-22}{2}
=-11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2}{2*1}=\frac{-18}{2}
=-9
$