-16t2+96t+200=0

Solution for -16t2+96t+200=0 equation:

-16t^2+96t+200=0

a = -16; b = 96; c = +200;
Δ = b2-4ac
Δ = 962-4·(-16)·200
Δ = 22016
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:

$\sqrt{\Delta}=\sqrt{22016}=\sqrt{256*86}=\sqrt{256}*\sqrt{86}=16\sqrt{86}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{86}}{2*-16}=\frac{-96-16\sqrt{86}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{86}}{2*-16}=\frac{-96+16\sqrt{86}}{-32}
$