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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-128t-0=0

We add all the numbers together, and all the variables

16t^2-128t=0

a = 16; b = -128; c = 0;
Δ = b2-4ac
Δ = -1282-4·16·0
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-128}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+128}{2*16}=\frac{256}{32}
=8
$