4.9t^2-4.9t+1=0

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

4.9t^2-4.9t+1=0

a = 4.9; b = -4.9; c = +1;
Δ = b2-4ac
Δ = -4.92-4·4.9·1
Δ = 4.41
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{-(-4.9)-\sqrt{4.41}}{2*4.9}=\frac{4.9-\sqrt{4.41}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4.9)+\sqrt{4.41}}{2*4.9}=\frac{4.9+\sqrt{4.41}}{9.8}
$