H(t)=V.t-4.9t^2

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Solution for H(t)=V.t-4.9t^2 equation: 



(H)=.H-4.9H^2

We move all terms to the left:

(H)-(.H-4.9H^2)=0

We get rid of parentheses

4.9H^2-.H+H=0

We add all the numbers together, and all the variables

4.9H^2=0

a = 4.9; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·4.9·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$H=\frac{-b}{2a}=\frac{0}{9.8}=0$

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