-4.905t^2+34.6t=0

Solution for -4.905t^2+34.6t=0 equation:

-4.905t^2+34.6t=0

a = -4.905; b = 34.6; c = 0;
Δ = b2-4ac
Δ = 34.62-4·(-4.905)·0
Δ = 1197.16
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{-(34.6)-\sqrt{1197.16}}{2*-4.905}=\frac{-34.6-\sqrt{1197.16}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34.6)+\sqrt{1197.16}}{2*-4.905}=\frac{-34.6+\sqrt{1197.16}}{-9.81}
$