128=-16t^2+96t+0

Solution for 128=-16t^2+96t+0 equation:

128=-16t^2+96t+0

We move all terms to the left:

128-(-16t^2+96t+0)=0

We get rid of parentheses

16t^2-96t-0+128=0

We add all the numbers together, and all the variables

16t^2-96t+128=0

a = 16; b = -96; c = +128;
Δ = b2-4ac
Δ = -962-4·16·128
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-32}{2*16}=\frac{64}{32}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+32}{2*16}=\frac{128}{32}
=4
$