-96=-16t^2+16t

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

-96=-16t^2+16t

We move all terms to the left:

-96-(-16t^2+16t)=0

We get rid of parentheses

16t^2-16t-96=0

a = 16; b = -16; c = -96;
Δ = b2-4ac
Δ = -162-4·16·(-96)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-80}{2*16}=\frac{-64}{32}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+80}{2*16}=\frac{96}{32}
=3
$