x^2-22x+80=0

Solution for x^2-22x+80=0 equation:

x^2-22x+80=0

a = 1; b = -22; c = +80;
Δ = b2-4ac
Δ = -222-4·1·80
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{41}}{2*1}=\frac{22-2\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{41}}{2*1}=\frac{22+2\sqrt{41}}{2}
$