0=80t-16t^2-96

Solution for 0=80t-16t^2-96 equation:

0=80t-16t^2-96

We move all terms to the left:

0-(80t-16t^2-96)=0

We add all the numbers together, and all the variables

-(80t-16t^2-96)=0

We get rid of parentheses

16t^2-80t+96=0

a = 16; b = -80; c = +96;
Δ = b2-4ac
Δ = -802-4·16·96
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-16}{2*16}=\frac{64}{32}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+16}{2*16}=\frac{96}{32}
=3
$