0=2.8+80t-16t^2

Solution for 0=2.8+80t-16t^2 equation:

0=2.8+80t-16t^2

We move all terms to the left:

0-(2.8+80t-16t^2)=0

We add all the numbers together, and all the variables

-(2.8+80t-16t^2)=0

We get rid of parentheses

16t^2-80t-2.8=0

a = 16; b = -80; c = -2.8;
Δ = b2-4ac
Δ = -802-4·16·(-2.8)
Δ = 6579.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-\sqrt{6579.2}}{2*16}=\frac{80-\sqrt{6579.2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+\sqrt{6579.2}}{2*16}=\frac{80+\sqrt{6579.2}}{32}
$