-16x^2+20x+0.5=0

Solution for -16x^2+20x+0.5=0 equation:

-16x^2+20x+0.5=0

a = -16; b = 20; c = +0.5;
Δ = b2-4ac
Δ = 202-4·(-16)·0.5
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{3}}{2*-16}=\frac{-20-12\sqrt{3}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{3}}{2*-16}=\frac{-20+12\sqrt{3}}{-32}
$