20t-4.905t^2=0

Solution for 20t-4.905t^2=0 equation:

20t-4.905t^2=0

a = -4.905; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-4.905)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-4.905}=\frac{-40}{-9.81}
=4+0.76/9.81
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-4.905}=\frac{0}{-9.81}
=0
$