4.905t^2-5t-1=0

Solution for 4.905t^2-5t-1=0 equation:

4.905t^2-5t-1=0

a = 4.905; b = -5; c = -1;
Δ = b2-4ac
Δ = -52-4·4.905·(-1)
Δ = 44.62
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{-(-5)-\sqrt{44.62}}{2*4.905}=\frac{5-\sqrt{44.62}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{44.62}}{2*4.905}=\frac{5+\sqrt{44.62}}{9.81}
$