0=-0.08t^2+1.6t+2

Solution for 0=-0.08t^2+1.6t+2 equation:

0=-0.08t^2+1.6t+2

We move all terms to the left:

0-(-0.08t^2+1.6t+2)=0

We add all the numbers together, and all the variables

-(-0.08t^2+1.6t+2)=0

We get rid of parentheses

0.08t^2-1.6t-2=0

a = 0.08; b = -1.6; c = -2;
Δ = b2-4ac
Δ = -1.62-4·0.08·(-2)
Δ = 3.2
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{-(-1.6)-\sqrt{3.2}}{2*0.08}=\frac{1.6-\sqrt{3.2}}{0.16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.6)+\sqrt{3.2}}{2*0.08}=\frac{1.6+\sqrt{3.2}}{0.16}
$