2x2+32x-122=0

Solution for 2x2+32x-122=0 equation:

2x^2+32x-122=0

a = 2; b = 32; c = -122;
Δ = b2-4ac
Δ = 322-4·2·(-122)
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-20\sqrt{5}}{2*2}=\frac{-32-20\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+20\sqrt{5}}{2*2}=\frac{-32+20\sqrt{5}}{4}
$