0=-16t^2+80t+.1

Solution for 0=-16t^2+80t+.1 equation:

0=-16t^2+80t+.1

We move all terms to the left:

0-(-16t^2+80t+.1)=0

We add all the numbers together, and all the variables

-(-16t^2+80t+.1)=0

We get rid of parentheses

16t^2-80t-.1=0

We add all the numbers together, and all the variables

16t^2-80t-0.1=0

a = 16; b = -80; c = -0.1;
Δ = b2-4ac
Δ = -802-4·16·(-0.1)
Δ = 6406.4
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{6406.4}}{2*16}=\frac{80-\sqrt{6406.4}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+\sqrt{6406.4}}{2*16}=\frac{80+\sqrt{6406.4}}{32}
$