Find all positive integers $x$ such that the product of all digits of $x$ is given by $x^2 - 10 \cdot x - 22.$
1. We start with the given equation:
   \[
   P(x) = x^2 - 10x - 22
   \]
   where \(P(x)\) is the product of the digits of \(x\).

2. Since \(x\) is a positive integer, the product of its digits must be non-negative. Therefore, we have:
   \[
   x^2 - 10x - 22 \geq 0
   \]

3. To solve the inequality \(x^2 - 10x - 22 \geq 0\), we first find the roots of the quadratic equation \(x^2 - 10x - 22 = 0\) using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \(a = 1\), \(b = -10\), and \(c = -22\). Plugging in these values, we get:
   \[
   x = \frac{10 \pm \sqrt{100 + 88}}{2} = \frac{10 \pm \sqrt{188}}{2} = \frac{10 \pm 2\sqrt{47}}{2} = 5 \pm \sqrt{47}
   \]

4. Since \(\sqrt{47}\) is approximately 6.86, the roots are approximately:
   \[
   x \approx 5 + 6.86 \approx 11.86 \quad \text{and} \quad x \approx 5 - 6.86 \approx -1.86
   \]
   Therefore, the inequality \(x^2 - 10x - 22 \geq 0\) holds for:
   \[
   x \leq 1 \quad \text{or} \quad x \geq 12
   \]

5. Since \(x\) is a positive integer, we discard \(x \leq 1\) and consider \(x \geq 12\).

6. Next, we consider the maximum possible product of the digits of \(x\). If \(x\) is a two-digit number, the maximum product of its digits is \(9 \times 9 = 81\). Therefore, we have:
   \[
   x^2 - 10x - 22 \leq 81
   \]

7. Solving the inequality \(x^2 - 10x - 22 \leq 81\), we get:
   \[
   x^2 - 10x - 103 \leq 0
   \]

8. To solve this quadratic inequality, we find the roots of the equation \(x^2 - 10x - 103 = 0\):
   \[
   x = \frac{10 \pm \sqrt{100 + 412}}{2} = \frac{10 \pm \sqrt{512}}{2} = \frac{10 \pm 16\sqrt{2}}{2} = 5 \pm 8\sqrt{2}
   \]

9. Since \(8\sqrt{2}\) is approximately 11.31, the roots are approximately:
   \[
   x \approx 5 + 11.31 \approx 16.31 \quad \text{and} \quad x \approx 5 - 11.31 \approx -6.31
   \]
   Therefore, the inequality \(x^2 - 10x - 103 \leq 0\) holds for:
   \[
   7 \leq x \leq 16
   \]

10. Combining the two ranges, \(x \geq 12\) and \(7 \leq x \leq 16\), we get:
    \[
    12 \leq x \leq 16
    \]

11. We now check each integer value in this range to see if it satisfies the original equation \(P(x) = x^2 - 10x - 22\):
    - For \(x = 12\):
      \[
      12^2 - 10 \cdot 12 - 22 = 144 - 120 - 22 = 2
      \]
      The product of the digits of 12 is \(1 \times 2 = 2\), which matches the right-hand side.
    - For \(x = 13\):
      \[
      13^2 - 10 \cdot 13 - 22 = 169 - 130 - 22 = 17
      \]
      The product of the digits of 13 is \(1 \times 3 = 3\), which does not match.
    - For \(x = 14\):
      \[
      14^2 - 10 \cdot 14 - 22 = 196 - 140 - 22 = 34
      \]
      The product of the digits of 14 is \(1 \times 4 = 4\), which does not match.
    - For \(x = 15\):
      \[
      15^2 - 10 \cdot 15 - 22 = 225 - 150 - 22 = 53
      \]
      The product of the digits of 15 is \(1 \times 5 = 5\), which does not match.
    - For \(x = 16\):
      \[
      16^2 - 10 \cdot 16 - 22 = 256 - 160 - 22 = 74
      \]
      The product of the digits of 16 is \(1 \times 6 = 6\), which does not match.

12. Therefore, the only integer \(x\) that satisfies the condition is \(x = 12\).

The final answer is \(\boxed{12}\).