0=-4.9t^2+5t+1.6

Solution for 0=-4.9t^2+5t+1.6 equation:

0=-4.9t^2+5t+1.6

We move all terms to the left:

0-(-4.9t^2+5t+1.6)=0

We add all the numbers together, and all the variables

-(-4.9t^2+5t+1.6)=0

We get rid of parentheses

4.9t^2-5t-1.6=0

a = 4.9; b = -5; c = -1.6;
Δ = b2-4ac
Δ = -52-4·4.9·(-1.6)
Δ = 56.36
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{-(-5)-\sqrt{56.36}}{2*4.9}=\frac{5-\sqrt{56.36}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{56.36}}{2*4.9}=\frac{5+\sqrt{56.36}}{9.8}
$