= Solution
Put $w=\sqrt{-14}$. The <ring of integers> is $\mathbb Z[w]$, with <discriminant> $-56$. The <Minkowski bound for ideal classes> for an imaginary quadratic field says that every <ideal class> has an integral representative of norm at most $(2/\pi)\sqrt{56}<5$. Therefore the <ideal class group> is generated by primes above two and three: norm-four representatives contribute only products of primes above two.
The prime two ramifies: $\mathfrak p_2=(2,w)$ has norm two and $\mathfrak p_2^2=(2)$. It is not principal, since an element of norm two would require $a^2+14b^2=2$, which has no integral solution. Thus $[\mathfrak p_2]$ has order two. Since $X^2+14=(X-1)(X+1)$ modulo three, three splits into $\mathfrak p_3=(3,w-1)$ and its conjugate, whose classes are inverses.
The element $2+w$ has norm $18$. It belongs to $\mathfrak p_3$ but not its conjugate, and its norm has exactly one factor two. Unique <prime ideal factorization> therefore gives
$$
(2+w)=\mathfrak p_2\mathfrak p_3^2.
$$
Hence $[\mathfrak p_3]^2=[\mathfrak p_2]^{-1}=[\mathfrak p_2]\ne1$. Since all generating classes are powers of $[\mathfrak p_3]$, \b[$\boxed{\operatorname{Cl}(\mathbb Q(\sqrt{-14}))\cong\mathbb Z/4\mathbb Z}$.]
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