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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-80t-80=0

a = 16; b = -80; c = -80;
Δ = b2-4ac
Δ = -802-4·16·(-80)
Δ = 11520
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:

$\sqrt{\Delta}=\sqrt{11520}=\sqrt{2304*5}=\sqrt{2304}*\sqrt{5}=48\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-48\sqrt{5}}{2*16}=\frac{80-48\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+48\sqrt{5}}{2*16}=\frac{80+48\sqrt{5}}{32}
$