0=-16x^2+40

Solution for 0=-16x^2+40 equation:

0=-16x^2+40

We move all terms to the left:

0-(-16x^2+40)=0

We add all the numbers together, and all the variables

-(-16x^2+40)=0

We get rid of parentheses

16x^2-40=0

a = 16; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·16·(-40)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*16}=\frac{0-16\sqrt{10}}{32}
=-\frac{16\sqrt{10}}{32}
=-\frac{\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*16}=\frac{0+16\sqrt{10}}{32}
=\frac{16\sqrt{10}}{32}
=\frac{\sqrt{10}}{2}
$