0=-16t^2+128t+80

Solution for 0=-16t^2+128t+80 equation:

0=-16t^2+128t+80

We move all terms to the left:

0-(-16t^2+128t+80)=0

We add all the numbers together, and all the variables

-(-16t^2+128t+80)=0

We get rid of parentheses

16t^2-128t-80=0

a = 16; b = -128; c = -80;
Δ = b2-4ac
Δ = -1282-4·16·(-80)
Δ = 21504
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:

$\sqrt{\Delta}=\sqrt{21504}=\sqrt{1024*21}=\sqrt{1024}*\sqrt{21}=32\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-32\sqrt{21}}{2*16}=\frac{128-32\sqrt{21}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+32\sqrt{21}}{2*16}=\frac{128+32\sqrt{21}}{32}
$