0=-16x^2+20x-5

Solution for 0=-16x^2+20x-5 equation:

0=-16x^2+20x-5

We move all terms to the left:

0-(-16x^2+20x-5)=0

We add all the numbers together, and all the variables

-(-16x^2+20x-5)=0

We get rid of parentheses

16x^2-20x+5=0

a = 16; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·16·5
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{5}}{2*16}=\frac{20-4\sqrt{5}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{5}}{2*16}=\frac{20+4\sqrt{5}}{32}
$