-16x^2+2=0

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

-16x^2+2=0

a = -16; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-16)·2
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{-(0)-8\sqrt{2}}{2*-16}=\frac{0-8\sqrt{2}}{-32}
=-\frac{8\sqrt{2}}{-32}
=-\frac{\sqrt{2}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-16}=\frac{0+8\sqrt{2}}{-32}
=\frac{8\sqrt{2}}{-32}
=\frac{\sqrt{2}}{-4}
$