2x^2+8x-22=0

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

2x^2+8x-22=0

a = 2; b = 8; c = -22;
Δ = b2-4ac
Δ = 82-4·2·(-22)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{15}}{2*2}=\frac{-8-4\sqrt{15}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{15}}{2*2}=\frac{-8+4\sqrt{15}}{4}
$