x^2-22x+22=0

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

x^2-22x+22=0

a = 1; b = -22; c = +22;
Δ = b2-4ac
Δ = -222-4·1·22
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-6\sqrt{11}}{2*1}=\frac{22-6\sqrt{11}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+6\sqrt{11}}{2*1}=\frac{22+6\sqrt{11}}{2}
$