-4x^2-20x+16=0

Solution for -4x^2-20x+16=0 equation:

-4x^2-20x+16=0

a = -4; b = -20; c = +16;
Δ = b2-4ac
Δ = -202-4·(-4)·16
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{41}}{2*-4}=\frac{20-4\sqrt{41}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{41}}{2*-4}=\frac{20+4\sqrt{41}}{-8}
$