2x^2+8x-122=0

Solution for 2x^2+8x-122=0 equation:

2x^2+8x-122=0

a = 2; b = 8; c = -122;
Δ = b2-4ac
Δ = 82-4·2·(-122)
Δ = 1040
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:

$\sqrt{\Delta}=\sqrt{1040}=\sqrt{16*65}=\sqrt{16}*\sqrt{65}=4\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{65}}{2*2}=\frac{-8-4\sqrt{65}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{65}}{2*2}=\frac{-8+4\sqrt{65}}{4}
$