-4.9t^2+19.33t+20=0

Solution for -4.9t^2+19.33t+20=0 equation:

-4.9t^2+19.33t+20=0

a = -4.9; b = 19.33; c = +20;
Δ = b2-4ac
Δ = 19.332-4·(-4.9)·20
Δ = 765.6489
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{-(19.33)-\sqrt{765.6489}}{2*-4.9}=\frac{-19.33-\sqrt{765.6489}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.33)+\sqrt{765.6489}}{2*-4.9}=\frac{-19.33+\sqrt{765.6489}}{-9.8}
$