r^2-50r+200=0

Solution for r^2-50r+200=0 equation:

r^2-50r+200=0

a = 1; b = -50; c = +200;
Δ = b2-4ac
Δ = -502-4·1·200
Δ = 1700
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:

$\sqrt{\Delta}=\sqrt{1700}=\sqrt{100*17}=\sqrt{100}*\sqrt{17}=10\sqrt{17}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{17}}{2*1}=\frac{50-10\sqrt{17}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{17}}{2*1}=\frac{50+10\sqrt{17}}{2}
$