-4x^2+32x+16=0

Solution for -4x^2+32x+16=0 equation:

-4x^2+32x+16=0

a = -4; b = 32; c = +16;
Δ = b2-4ac
Δ = 322-4·(-4)·16
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{-(32)-16\sqrt{5}}{2*-4}=\frac{-32-16\sqrt{5}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{5}}{2*-4}=\frac{-32+16\sqrt{5}}{-8}
$