0=-16t^2+t+80

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

0=-16t^2+t+80

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-t-80=0

We add all the numbers together, and all the variables

16t^2-1t-80=0

a = 16; b = -1; c = -80;
Δ = b2-4ac
Δ = -12-4·16·(-80)
Δ = 5121
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:

$\sqrt{\Delta}=\sqrt{5121}=\sqrt{9*569}=\sqrt{9}*\sqrt{569}=3\sqrt{569}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3\sqrt{569}}{2*16}=\frac{1-3\sqrt{569}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3\sqrt{569}}{2*16}=\frac{1+3\sqrt{569}}{32}
$