Find all natural numbers $x$ such that the product of their digits (in decimal notation) is equal to
$$ x^2 - 10x-22. $$
Let us denote the digits of the number $x$ from left to right as $a_n, a_{n-1}, \dots, a_0$. First, notice that
$$ 0\leq \prod_{i = 0}^n a_i = a_n\prod_{i=0}^{n-1}a_i < a_i\cdot 10^{n}\leq x. $$
Substituting $\prod_{i = 0}^n a_i = x^2 - 10x - 22$ we obtain
$$ x^2 - 11x - 22 \leq 0 \leq x^2 - 10x - 22. $$
Solving these equations and using that $x\geq 0$ we obtain
$$ \begin{align*} 5 + \sqrt{47}\leq & x \leq \frac{11 + \sqrt{209}}{2}\\ 11 = 5 + \sqrt{36} < & x < \frac{11 + 15}{2} = 13. \end{align*} $$
Thus the only feasible answer would be $x = 12$. Let us check it:
$$
12^2 - 10\cdot 12 - 22 = 12\cdot 2 - 22 = 2 = 1\cdot 2,
$$
so $x = 12$ is the solution.
Shoutout to Matvei Zhukov (Finland), Felix, (Germany/Würzburg), Olivier Massicot (Champaign, IL), jc (New York), Michael Vincent (Washington), and B Sashi Kanth (Hyderabad, India) for submitting solutions to this challenge problem.