0=80-4t-16t2

Solution for 0=80-4t-16t2 equation:

0=80-4t-16t^2

We move all terms to the left:

0-(80-4t-16t^2)=0

We add all the numbers together, and all the variables

-(80-4t-16t^2)=0

We get rid of parentheses

16t^2+4t-80=0

a = 16; b = 4; c = -80;
Δ = b2-4ac
Δ = 42-4·16·(-80)
Δ = 5136
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:

$\sqrt{\Delta}=\sqrt{5136}=\sqrt{16*321}=\sqrt{16}*\sqrt{321}=4\sqrt{321}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{321}}{2*16}=\frac{-4-4\sqrt{321}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{321}}{2*16}=\frac{-4+4\sqrt{321}}{32}
$