2x^2+12x-6=32

Solution for 2x^2+12x-6=32 equation:

2x^2+12x-6=32

We move all terms to the left:

2x^2+12x-6-(32)=0

We add all the numbers together, and all the variables

2x^2+12x-38=0

a = 2; b = 12; c = -38;
Δ = b2-4ac
Δ = 122-4·2·(-38)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{7}}{2*2}=\frac{-12-8\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{7}}{2*2}=\frac{-12+8\sqrt{7}}{4}
$