2x^2+22x-26=0

Solution for 2x^2+22x-26=0 equation:

2x^2+22x-26=0

a = 2; b = 22; c = -26;
Δ = b2-4ac
Δ = 222-4·2·(-26)
Δ = 692
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:

$\sqrt{\Delta}=\sqrt{692}=\sqrt{4*173}=\sqrt{4}*\sqrt{173}=2\sqrt{173}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{173}}{2*2}=\frac{-22-2\sqrt{173}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{173}}{2*2}=\frac{-22+2\sqrt{173}}{4}
$