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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-96t-128=0

a = 16; b = -96; c = -128;
Δ = b2-4ac
Δ = -962-4·16·(-128)
Δ = 17408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{17408}=\sqrt{1024*17}=\sqrt{1024}*\sqrt{17}=32\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-32\sqrt{17}}{2*16}=\frac{96-32\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+32\sqrt{17}}{2*16}=\frac{96+32\sqrt{17}}{32}
$