-16x^2+32x+160=0

Solution for -16x^2+32x+160=0 equation:

-16x^2+32x+160=0

a = -16; b = 32; c = +160;
Δ = b2-4ac
Δ = 322-4·(-16)·160
Δ = 11264
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:

$\sqrt{\Delta}=\sqrt{11264}=\sqrt{1024*11}=\sqrt{1024}*\sqrt{11}=32\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32\sqrt{11}}{2*-16}=\frac{-32-32\sqrt{11}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32\sqrt{11}}{2*-16}=\frac{-32+32\sqrt{11}}{-32}
$