How to calculate allele frequencies - PTC tasting problem solving

In 1908, two scientists, Godfrey H. Hardy, an English mathematician, and Wilhelm Weinberg, a German physician, independently worked out a mathematical relationship that related genotypes to allele frequencies.05
Hardy
Godfrey H. Hardy Weinberg
Wilhelm Weinberg
Their mathematical concept, called the Hardy-Weinberg principle, is a crucial concept in population genetics. It predicts how gene frequencies will be inherited from generation to generation given a specific set of assumptions.06 The Hardy-Weinberg principle states that in a large randomly breeding population, allelic frequencies will remain the same from generation to generation assuming that there is no mutation, gene migration, selection or genetic drift. This principle is important because it gives biologists a standard from which to measure changes in allele frequency in a population.
To illustrate how the Hardy-Weinberg principle works, let us consider the MN blood group. Humans inherit either the M or the N antigen which is determined by two different alleles at the same gene locus. If we let the frequency of allele M=p and the frequency of the other allele N=q, then the next generation's genotypes will occur as follows:
Frequency of MM genotype = p2
Frequency of MN genotype = 2pq
Frequency of NN genotype = q2
We can take a sample of the population and count the number of people with each genotype. For example, a sample of 5000 from Forensic Town, USA, has:
1460 individuals of type MM, that is 1460/5000 or 29.2%
2550 of type MN, that is 2550/5000 or 51%
990 of type NN, that is 990/5000 or 19.8%
If we apply the Hardy-Weinberg equation (p2 + 2pq + q2 = 1) we can calculate the allele frequencies as:
Frequency of M = p2 + 0.5 (2pq) = 0.292 + (0.5 x 0.51) = 0.547
Frequency of N = q = 1 - p = 1 - 0.547 = 0.453
We can now calculate our expected genotype frequencies:
MM = p2 = 0.5472 = 0.299, or 1496 individuals in the sample
MN = 2pq = 2 x 0.547 x 0.453 = 0.496, or 2478 individuals
NN = q2 = 0.4532 = 0.205, or 1026 individuals