4.6t-4.9t^2=0

Solution for 4.6t-4.9t^2=0 equation:

4.6t-4.9t^2=0

a = -4.9; b = 4.6; c = 0;
Δ = b2-4ac
Δ = 4.62-4·(-4.9)·0
Δ = 21.16
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{-(4.6)-\sqrt{21.16}}{2*-4.9}=\frac{-4.6-\sqrt{21.16}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.6)+\sqrt{21.16}}{2*-4.9}=\frac{-4.6+\sqrt{21.16}}{-9.8}
$