0=-16x^2+40x+20

Solution for 0=-16x^2+40x+20 equation:

0=-16x^2+40x+20

We move all terms to the left:

0-(-16x^2+40x+20)=0

We add all the numbers together, and all the variables

-(-16x^2+40x+20)=0

We get rid of parentheses

16x^2-40x-20=0

a = 16; b = -40; c = -20;
Δ = b2-4ac
Δ = -402-4·16·(-20)
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24\sqrt{5}}{2*16}=\frac{40-24\sqrt{5}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24\sqrt{5}}{2*16}=\frac{40+24\sqrt{5}}{32}
$