-4x^2=16x+8=0

Solution for -4x^2=16x+8=0 equation:

-4x^2=16x+8=0

We move all terms to the left:

-4x^2-(16x+8)=0

We get rid of parentheses

-4x^2-16x-8=0

a = -4; b = -16; c = -8;
Δ = b2-4ac
Δ = -162-4·(-4)·(-8)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{2}}{2*-4}=\frac{16-8\sqrt{2}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{2}}{2*-4}=\frac{16+8\sqrt{2}}{-8}
$