3=16t+4.9t^2

Solution for 3=16t+4.9t^2 equation:

3=16t+4.9t^2

We move all terms to the left:

3-(16t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-16t+3=0

a = -4.9; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·(-4.9)·3
Δ = 314.8
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{314.8}}{2*-4.9}=\frac{16-\sqrt{314.8}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+\sqrt{314.8}}{2*-4.9}=\frac{16+\sqrt{314.8}}{-9.8}
$