0=-16t^2+96t+80

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

0=-16t^2+96t+80

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-96t-80=0

a = 16; b = -96; c = -80;
Δ = b2-4ac
Δ = -962-4·16·(-80)
Δ = 14336
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:

$\sqrt{\Delta}=\sqrt{14336}=\sqrt{1024*14}=\sqrt{1024}*\sqrt{14}=32\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-32\sqrt{14}}{2*16}=\frac{96-32\sqrt{14}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+32\sqrt{14}}{2*16}=\frac{96+32\sqrt{14}}{32}
$