-8x^2+40=0

Solution for -8x^2+40=0 equation:

-8x^2+40=0

a = -8; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-8)·40
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*-8}=\frac{0-16\sqrt{5}}{-16}
=-\frac{16\sqrt{5}}{-16}
=-\frac{\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*-8}=\frac{0+16\sqrt{5}}{-16}
=\frac{16\sqrt{5}}{-16}
=\frac{\sqrt{5}}{-1}
$