2x^2+32x-8=14

Solution for 2x^2+32x-8=14 equation:

2x^2+32x-8=14

We move all terms to the left:

2x^2+32x-8-(14)=0

We add all the numbers together, and all the variables

2x^2+32x-22=0

a = 2; b = 32; c = -22;
Δ = b2-4ac
Δ = 322-4·2·(-22)
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-20\sqrt{3}}{2*2}=\frac{-32-20\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+20\sqrt{3}}{2*2}=\frac{-32+20\sqrt{3}}{4}
$