-4.9t^2+19.6t+0.3=0

Solution for -4.9t^2+19.6t+0.3=0 equation:

-4.9t^2+19.6t+0.3=0

a = -4.9; b = 19.6; c = +0.3;
Δ = b2-4ac
Δ = 19.62-4·(-4.9)·0.3
Δ = 390.04
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{-(19.6)-\sqrt{390.04}}{2*-4.9}=\frac{-19.6-\sqrt{390.04}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.6)+\sqrt{390.04}}{2*-4.9}=\frac{-19.6+\sqrt{390.04}}{-9.8}
$