22x^2-22x+4=0

Solution for 22x^2-22x+4=0 equation:

22x^2-22x+4=0

a = 22; b = -22; c = +4;
Δ = b2-4ac
Δ = -222-4·22·4
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{33}}{2*22}=\frac{22-2\sqrt{33}}{44}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{33}}{2*22}=\frac{22+2\sqrt{33}}{44}
$