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

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

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

a = -16; b = 96; c = +96;
Δ = b2-4ac
Δ = 962-4·(-16)·96
Δ = 15360
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:

$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-32\sqrt{15}}{2*-16}=\frac{-96-32\sqrt{15}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+32\sqrt{15}}{2*-16}=\frac{-96+32\sqrt{15}}{-32}
$