1x^2+22x-8=0

Solution for 1x^2+22x-8=0 equation:

1x^2+22x-8=0

We add all the numbers together, and all the variables

x^2+22x-8=0

a = 1; b = 22; c = -8;
Δ = b2-4ac
Δ = 222-4·1·(-8)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{129}}{2*1}=\frac{-22-2\sqrt{129}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{129}}{2*1}=\frac{-22+2\sqrt{129}}{2}
$