19.6t-4.9t^2+103=0

Solution for 19.6t-4.9t^2+103=0 equation:

19.6t-4.9t^2+103=0

a = -4.9; b = 19.6; c = +103;
Δ = b2-4ac
Δ = 19.62-4·(-4.9)·103
Δ = 2402.96
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{2402.96}}{2*-4.9}=\frac{-19.6-\sqrt{2402.96}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.6)+\sqrt{2402.96}}{2*-4.9}=\frac{-19.6+\sqrt{2402.96}}{-9.8}
$