0=-4.9t^2+16t+2

Solution for 0=-4.9t^2+16t+2 equation:

0=-4.9t^2+16t+2

We move all terms to the left:

0-(-4.9t^2+16t+2)=0

We add all the numbers together, and all the variables

-(-4.9t^2+16t+2)=0

We get rid of parentheses

4.9t^2-16t-2=0

a = 4.9; b = -16; c = -2;
Δ = b2-4ac
Δ = -162-4·4.9·(-2)
Δ = 295.2
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{-(-16)-\sqrt{295.2}}{2*4.9}=\frac{16-\sqrt{295.2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+\sqrt{295.2}}{2*4.9}=\frac{16+\sqrt{295.2}}{9.8}
$