2x^2+8x-23=0

Solution for 2x^2+8x-23=0 equation:

2x^2+8x-23=0

a = 2; b = 8; c = -23;
Δ = b2-4ac
Δ = 82-4·2·(-23)
Δ = 248
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:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{62}}{2*2}=\frac{-8-2\sqrt{62}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{62}}{2*2}=\frac{-8+2\sqrt{62}}{4}
$