879. Solve the equation in natural numbers

$$
x!+12=y^{2}
$$
$\triangle$ For $x>4$ the number $x!$ ends with the digit 0, which means the number $x!+12$ ends with the digit 2. However, the square of a natural number cannot end with the digit 2, so there are no solutions to the equation for $x>4$.

It remains to check the values $x=1,2,3,$ and 4. A solution is obtained only for $x=4$:

$$
4!+12=y^{2}, \quad y^{2}=24+12=36, \quad y=6
$$

Answer: $(4 ; 6)$.