= Solution
Put $s=\sqrt{30}$. Since $30\equiv2\pmod4$, the <ring of integers> is $\mathbb Z[s]$ and its <field discriminant> is $120$. All three primes ramify:
$$
\boxed{(p)=\mathfrak p_p^2,\qquad\mathfrak p_p=(p,s),\qquad N\mathfrak p_p=p,\quad p=2,3,5.}
$$
For example $\mathfrak p_p^2\subseteq(p)$ and both ideals have norm $p^2$, proving equality. The <Minkowski bound for ideal classes> is $\sqrt{120}/2=\sqrt{30}<6$. Thus the <ideal class group> is generated by the classes of these prime ideals; an ideal of norm four contributes only the already principal $\mathfrak p_2^2$.
The elements $6+s$ and $5+s$ have norms $6$ and $-5$. Their ideal factorizations give $\mathfrak p_2\mathfrak p_3=(6+s)$ and $\mathfrak p_5=(5+s)$. This also explains the printed hint: $N((6+s)/2)=3/2$, a fraction, not the integer $32$. Therefore $[\mathfrak p_3]=[\mathfrak p_2]$ and $[\mathfrak p_5]=1$, leaving at most two ideal classes. The class of $\mathfrak p_2$ is nontrivial: a principal generator would have norm $\pm2$, but $x^2-30y^2=\pm2$ would require $x^2\equiv2$ or $3\pmod5$, impossible. Hence
$$
\boxed{\operatorname{Cl}(\mathbb Q(\sqrt{30}))\cong\mathbb Z/2\mathbb Z.}
$$
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