q^2-80q+500=0

Solution for q^2-80q+500=0 equation:

q^2-80q+500=0

a = 1; b = -80; c = +500;
Δ = b2-4ac
Δ = -802-4·1·500
Δ = 4400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4400}=\sqrt{400*11}=\sqrt{400}*\sqrt{11}=20\sqrt{11}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-20\sqrt{11}}{2*1}=\frac{80-20\sqrt{11}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+20\sqrt{11}}{2*1}=\frac{80+20\sqrt{11}}{2}
$