-8x^2+10=0

Solution for -8x^2+10=0 equation:

-8x^2+10=0

a = -8; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-8)·10
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*-8}=\frac{0-8\sqrt{5}}{-16}
=-\frac{8\sqrt{5}}{-16}
=-\frac{\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*-8}=\frac{0+8\sqrt{5}}{-16}
=\frac{8\sqrt{5}}{-16}
=\frac{\sqrt{5}}{-2}
$