-16t^2+96t+128=0

Solution for -16t^2+96t+128=0 equation:

-16t^2+96t+128=0

a = -16; b = 96; c = +128;
Δ = b2-4ac
Δ = 962-4·(-16)·128
Δ = 17408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{17408}=\sqrt{1024*17}=\sqrt{1024}*\sqrt{17}=32\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-32\sqrt{17}}{2*-16}=\frac{-96-32\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+32\sqrt{17}}{2*-16}=\frac{-96+32\sqrt{17}}{-32}
$