2x^2+16x=16=0

Solution for 2x^2+16x=16=0 equation:

2x^2+16x=16=0

We move all terms to the left:

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

a = 2; b = 16; c = -16;
Δ = b2-4ac
Δ = 162-4·2·(-16)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{6}}{2*2}=\frac{-16-8\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{6}}{2*2}=\frac{-16+8\sqrt{6}}{4}
$