-16x^2-40x+8=0

Solution for -16x^2-40x+8=0 equation:

-16x^2-40x+8=0

a = -16; b = -40; c = +8;
Δ = b2-4ac
Δ = -402-4·(-16)·8
Δ = 2112
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:

$\sqrt{\Delta}=\sqrt{2112}=\sqrt{64*33}=\sqrt{64}*\sqrt{33}=8\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{33}}{2*-16}=\frac{40-8\sqrt{33}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{33}}{2*-16}=\frac{40+8\sqrt{33}}{-32}
$