Product of digits Find all natural numbers $x$ ($x$ in base $10$) so that the product of its digits is $x^2 - 10x - 22$.
Here is what I did so far:
I took two cases. 
The first case was considering one or more digits(except the first one) being $0$. In this case $x$ wasn't a natural number. 
The second case was considering no digit being $0$. Because $x$ is a natural number we must have the product of its digits bigger than $0$. After solving $x^2 - 10x - 22 > 0$ I got $x$ being a natural number bigger than $11$. 
At this point I got stucked. I know the answer is $12$ but I don't know how to get it.
 A: $\lg(x)$ here denotes $\log_{10}(x)$. 
$x^2-10x-12\neq 0$, since $x$ is an integer.   
$x^2-10x-22 > 0\iff x\ge 12$.  
As suggested by Daniel Fischer in the comments: 
$$x^2-10x-22\le 9^{d}=9^{\lfloor \lg x \rfloor+1}$$  
$$\iff \lg(x^2-10x-22)\le \lg(9)\left(\lfloor\lg x\rfloor +1\right)< \lfloor \lg x\rfloor+1\le \lg x+1=\lg (10x)$$
$$\implies x^2-10x-22< 10x\iff x\in (10-\sqrt{122},10+\sqrt{122})$$   
$\iff x\in [12,21]$. Then $0\le x^2-10x-22\le 1\cdot 9\implies x=12$. After checking, $12$ works.
A: All digit $a$ is equal to the product of its digits then none digit is solution because of $a^2-10a-22$. On the other hand this trinomial is equal to $(x - 5)^2 – 47$ which shows your solution $x = 12$ (and discards the numbers 10 and 11). 
For numbers of two digits, other than the 12 above we have $(x - 5)^2$ – 47 = ab$ \implies$ $(x - 5)^2 – 47\leq81$ $\implies(x-5)^2 = 9, 16, 25$  i.e. $x = 14, 21, 30$. But $9^2 – 47\neq{1\cdot4}$, ${16}^2 – 47\neq{2\cdot1}$ and ${25}^2 – 47\neq{3\cdot0}$. Thus the only solution of two digits is the 12 above.
For numbers of three digits we would have $(x - 5)^2 – 47\leq729=9^3$ but $(x - 5)^2\geq(95)^2$ which is not possible.
The only solution is $x = 12$
