-4.905t^2+5t=0

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

-4.905t^2+5t=0

a = -4.905; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-4.905)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-4.905}=\frac{-10}{-9.81}
=1+0.19/9.81
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-4.905}=\frac{0}{-9.81}
=0
$