-16x^2+16x+192=0

Solution for -16x^2+16x+192=0 equation:

-16x^2+16x+192=0

a = -16; b = 16; c = +192;
Δ = b2-4ac
Δ = 162-4·(-16)·192
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-112}{2*-16}=\frac{-128}{-32}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+112}{2*-16}=\frac{96}{-32}
=-3
$