-16t2+96t=0

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

-16t^2+96t=0

a = -16; b = 96; c = 0;
Δ = b2-4ac
Δ = 962-4·(-16)·0
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-96}{2*-16}=\frac{-192}{-32}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+96}{2*-16}=\frac{0}{-32}
=0
$