= Solution
Here $\mathcal O_K=\mathbb Z[s]$ with $s=\sqrt{-14}$ and <field discriminant> $-56$. The imaginary-quadratic <Minkowski bound for ideal classes> is $(2/\pi)\sqrt{56}<5$, so each <ideal class> has an integral representative of <ideal norm> at most four. Representatives therefore use <prime ideals> above two and three. Put
$$
\mathfrak p=(2,s),\qquad\mathfrak q=(3,s+1).
$$
The <prime> two ramifies, with $\mathfrak p^2=(2)$, and three splits into $\mathfrak q\overline{\mathfrak q}=(3)$. No element has <field norm> two, because $N(a+bs)=a^2+14b^2$; thus $[\mathfrak p]$ is nontrivial of order two. Norm-four <ideals> represent $[\mathfrak p]^2=1$ and add no generator, so the class group is generated by $[\mathfrak p],[\mathfrak q]$.
The element $2-s$ has <field norm> 18. It belongs to $\mathfrak p$ and $\mathfrak q$, but not $\overline{\mathfrak q}$, since reduction there gives $s\equiv1\pmod3$. Its <ideal> therefore factors as
$$
(2-s)=\mathfrak p\mathfrak q^2.
$$
The exponents follow from the <ideal norm> 18, or from the root $s\equiv2\pmod9$ defining $\mathfrak q^2$. Hence $[\mathfrak q]^2=[\mathfrak p]$, so $[\mathfrak q]$ has order four and already generates all possible classes. Therefore
$$
\boxed{\operatorname{Cl}(\mathbb Q(\sqrt{-14}))\cong\mathbb Z/4\mathbb Z.}
$$
As a separate check, the reduced forms of <discriminant of a binary quadratic form> $-56$ are $[1,0,14],[2,0,7],[3,2,5],[3,-2,5]$, giving exactly four classes.
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