-16t^2+96t-100=0

Solution for -16t^2+96t-100=0 equation:

-16t^2+96t-100=0

a = -16; b = 96; c = -100;
Δ = b2-4ac
Δ = 962-4·(-16)·(-100)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{11}}{2*-16}=\frac{-96-16\sqrt{11}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{11}}{2*-16}=\frac{-96+16\sqrt{11}}{-32}
$