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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-96t=0

a = 16; b = -96; c = 0;
Δ = b2-4ac
Δ = -962-4·16·0
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-96}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+96}{2*16}=\frac{192}{32}
=6
$