r^2-40r+100=0

Solution for r^2-40r+100=0 equation:

r^2-40r+100=0

a = 1; b = -40; c = +100;
Δ = b2-4ac
Δ = -402-4·1·100
Δ = 1200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20\sqrt{3}}{2*1}=\frac{40-20\sqrt{3}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20\sqrt{3}}{2*1}=\frac{40+20\sqrt{3}}{2}
$