-16a^2+8=0

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

-16a^2+8=0

a = -16; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-16)·8
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*-16}=\frac{0-16\sqrt{2}}{-32}
=-\frac{16\sqrt{2}}{-32}
=-\frac{\sqrt{2}}{-2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*-16}=\frac{0+16\sqrt{2}}{-32}
=\frac{16\sqrt{2}}{-32}
=\frac{\sqrt{2}}{-2}
$