2x^2-22x+8=0

Solution for 2x^2-22x+8=0 equation:

2x^2-22x+8=0

a = 2; b = -22; c = +8;
Δ = b2-4ac
Δ = -222-4·2·8
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{105}}{2*2}=\frac{22-2\sqrt{105}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{105}}{2*2}=\frac{22+2\sqrt{105}}{4}
$