-16x2+16x+96=0

Solution for -16x2+16x+96=0 equation:

-16x^2+16x+96=0

a = -16; b = 16; c = +96;
Δ = b2-4ac
Δ = 162-4·(-16)·96
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-80}{2*-16}=\frac{-96}{-32}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+80}{2*-16}=\frac{64}{-32}
=-2
$