In this case, we have a double-cross between a dominant homozygous for both traits and a recessive heterozygous. According to the exercise we have red for dominant (R) white for recessive (w), tal dominant (T), and short recessive (t). So the parents are as follows
Red tall homozygous (RRTT)
White tal heterozygous (wwTt)
We obtain gametes for each parent

Now we obtain the punnet square
Now we obtain frequencies
Genotypes:
Red heterozygous tall homozygous RwTT: 8/16= 50%
Red heterozygous tall heterozygous RwTt: 8/16= 50%
Phenotypes:
Red tall 16/16=100%
Now the gardener would never obtain red short flowers because there is none recessive homozygous for height trait.