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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2+32t=0

a = 16; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·16·0
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{-(32)-32}{2*16}=\frac{-64}{32}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*16}=\frac{0}{32}
=0
$