2x^2+16x-32=6

Solution for 2x^2+16x-32=6 equation:

2x^2+16x-32=6

We move all terms to the left:

2x^2+16x-32-(6)=0

We add all the numbers together, and all the variables

2x^2+16x-38=0

a = 2; b = 16; c = -38;
Δ = b2-4ac
Δ = 162-4·2·(-38)
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{35}}{2*2}=\frac{-16-4\sqrt{35}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{35}}{2*2}=\frac{-16+4\sqrt{35}}{4}
$