4.9t^2-19.6t-103=0

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

4.9t^2-19.6t-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}
$