Find all real numbers $x$ for which there exists a triangle $A B C$ with circumradius 2, such that $\angle A B C \geq 90^{\circ}$, and
$$
x^{4}+a x^{3}+b x^{2}+c x+1=0
$$
where $a=B C, b=C A, c=A B$.
The answer is $x=-\frac{1}{2}(\sqrt{6} \pm \sqrt{2})$.
We prove this the only possible answer. Evidently $x0$ implies
$$
\{a, c\}=\{\sqrt{6} \pm \sqrt{2}\} .
$$

This gives the $x$ values claimed above; by taking $a, b, c$ as deduced here, we find they work too.
Remark. Note that by perturbing $\triangle A B C$ slightly, we see a priori that the set of possible $x$ should consist of unions of intervals (possibly trivial). So it makes sense to try inequalities no matter what.