-4.905t^2+25.54t=0

Solution for -4.905t^2+25.54t=0 equation:

-4.905t^2+25.54t=0

a = -4.905; b = 25.54; c = 0;
Δ = b2-4ac
Δ = 25.542-4·(-4.905)·0
Δ = 652.2916
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{-(25.54)-\sqrt{652.2916}}{2*-4.905}=\frac{-25.54-\sqrt{652.2916}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25.54)+\sqrt{652.2916}}{2*-4.905}=\frac{-25.54+\sqrt{652.2916}}{-9.81}
$