-16t^2+396t+100=0

Solution for -16t^2+396t+100=0 equation:

-16t^2+396t+100=0

a = -16; b = 396; c = +100;
Δ = b2-4ac
Δ = 3962-4·(-16)·100
Δ = 163216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{163216}=404$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(396)-404}{2*-16}=\frac{-800}{-32}
=+25
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(396)+404}{2*-16}=\frac{8}{-32}
=-1/4
$