4.9t^2-2.48t-1=0

Solution for 4.9t^2-2.48t-1=0 equation:

4.9t^2-2.48t-1=0

a = 4.9; b = -2.48; c = -1;
Δ = b2-4ac
Δ = -2.482-4·4.9·(-1)
Δ = 25.7504
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{-(-2.48)-\sqrt{25.7504}}{2*4.9}=\frac{2.48-\sqrt{25.7504}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.48)+\sqrt{25.7504}}{2*4.9}=\frac{2.48+\sqrt{25.7504}}{9.8}
$