r^2-50r+450=0

Solution for r^2-50r+450=0 equation:

r^2-50r+450=0

a = 1; b = -50; c = +450;
Δ = b2-4ac
Δ = -502-4·1·450
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{7}}{2*1}=\frac{50-10\sqrt{7}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{7}}{2*1}=\frac{50+10\sqrt{7}}{2}
$