d^2-20d+100=1

Solution for d^2-20d+100=1 equation:

d^2-20d+100=1

We move all terms to the left:

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

We add all the numbers together, and all the variables

d^2-20d+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:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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