What is the answer to this Hardy-Weinberg problem?

Hardy-Weinberg equation is p^2 + 2pq + q^2 = 1

p is representative of the frequency of the dominant allele of a gene and q the recessive. The equation shows the frequency of different combinations of these two alleles for a gene with two alleles in HW equilibrium. p^2 is homozygous dominant, q^2 homozygous recessive and 2pq is heterozygotes.

The frequencies are calculated by putting the number of a category over the total plants (440).

Therefore you should be able to calculate each frequency and plug it into the formula to get your p and q value

ZZ is homozygous dominant so it's p^2, Zz is 2pq and zz is q^2

 
For an example p^2 (ZZ) is 100/440 = .23

The square root of .23 is .48 and therefor p=.48

since p + q = 1
q must equal .52