2x^2+32x-117=0

Solution for 2x^2+32x-117=0 equation:

2x^2+32x-117=0

a = 2; b = 32; c = -117;
Δ = b2-4ac
Δ = 322-4·2·(-117)
Δ = 1960
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:

$\sqrt{\Delta}=\sqrt{1960}=\sqrt{196*10}=\sqrt{196}*\sqrt{10}=14\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-14\sqrt{10}}{2*2}=\frac{-32-14\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+14\sqrt{10}}{2*2}=\frac{-32+14\sqrt{10}}{4}
$