0.003=-0.3t-4.9t2

Solution for 0.003=-0.3t-4.9t2 equation:

0.003=-0.3t-4.9t^2

We move all terms to the left:

0.003-(-0.3t-4.9t^2)=0

We get rid of parentheses

4.9t^2+0.3t+0.003=0

a = 4.9; b = 0.3; c = +0.003;
Δ = b2-4ac
Δ = 0.32-4·4.9·0.003
Δ = 0.0312
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{-(0.3)-\sqrt{0.0312}}{2*4.9}=\frac{-0.3-\sqrt{0.0312}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.0312}}{2*4.9}=\frac{-0.3+\sqrt{0.0312}}{9.8}
$