n^2-200*n+200=0

Solution for n^2-200*n+200=0 equation:

n^2-200n+200=0

a = 1; b = -200; c = +200;
Δ = b2-4ac
Δ = -2002-4·1·200
Δ = 39200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{39200}=\sqrt{19600*2}=\sqrt{19600}*\sqrt{2}=140\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-200)-140\sqrt{2}}{2*1}=\frac{200-140\sqrt{2}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-200)+140\sqrt{2}}{2*1}=\frac{200+140\sqrt{2}}{2}
$