-16t^2-5t+100=0

Solution for -16t^2-5t+100=0 equation:

-16t^2-5t+100=0

a = -16; b = -5; c = +100;
Δ = b2-4ac
Δ = -52-4·(-16)·100
Δ = 6425
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:

$\sqrt{\Delta}=\sqrt{6425}=\sqrt{25*257}=\sqrt{25}*\sqrt{257}=5\sqrt{257}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{257}}{2*-16}=\frac{5-5\sqrt{257}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{257}}{2*-16}=\frac{5+5\sqrt{257}}{-32}
$