calculate the number of heterozygotes
Hardy-Weinberg law gives the frequencies of genotypes from allele frequencies: $p^2 + 2pq + q^2$

Because the allele "T" is dominant, the frequency you're directly measuring through the experiment is the addition of $p^2$ and $2pq$.

$$p^2 + 2pq = 150/215 ≈ 0.6976$$

solving for $p$ (and replacing $q$ by $1-p$) gives:

$$p = 1 - \sqrt{\frac{13}{42}} ≈ 0.45 $$

The frequency of heterozygotes is given by:

$$2p(1-p) ≈ 0.4950 $$

And the number of heterozygotes in a population of 215 individuals is given by:

$$215 \cdot 0.4950 = 106.43$$

That sounds very alike what you wrote in the comments!