0=-64t^2+64t+80

Solution for 0=-64t^2+64t+80 equation:

0=-64t^2+64t+80

We move all terms to the left:

0-(-64t^2+64t+80)=0

We add all the numbers together, and all the variables

-(-64t^2+64t+80)=0

We get rid of parentheses

64t^2-64t-80=0

a = 64; b = -64; c = -80;
Δ = b2-4ac
Δ = -642-4·64·(-80)
Δ = 24576
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:

$\sqrt{\Delta}=\sqrt{24576}=\sqrt{4096*6}=\sqrt{4096}*\sqrt{6}=64\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-64\sqrt{6}}{2*64}=\frac{64-64\sqrt{6}}{128}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+64\sqrt{6}}{2*64}=\frac{64+64\sqrt{6}}{128}
$