-4x^2+40x-20=0

Solution for -4x^2+40x-20=0 equation:

-4x^2+40x-20=0

a = -4; b = 40; c = -20;
Δ = b2-4ac
Δ = 402-4·(-4)·(-20)
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{-(40)-16\sqrt{5}}{2*-4}=\frac{-40-16\sqrt{5}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{5}}{2*-4}=\frac{-40+16\sqrt{5}}{-8}
$