The equation \(X^2 + 19X - X! = 0\) is a quadratic equation combined with a factorial term. To solve it, we'll first need to check integer values for \(X\) since the factorial function \(X!\) is only defined for non-negative integers.

### Steps:
1. Rewrite the equation as:
   \[
   X^2 + 19X = X!
   \]

2. Check possible integer values for \(X\), as \(X!\) grows very quickly.

### Test integer values:

- **For \(X = 0\):**
  \[
  0^2 + 19(0) = 0! \Rightarrow 0 = 1 \quad (\text{False})
  \]
  
- **For \(X = 1\):**
  \[
  1^2 + 19(1) = 1! \Rightarrow 1 + 19 = 1 \quad (\text{False})
  \]

- **For \(X = 2\):**
  \[
  2^2 + 19(2) = 2! \Rightarrow 4 + 38 = 2 \quad (\text{False})
  \]

- **For \(X = 3\):**
  \[
  3^2 + 19(3) = 3! \Rightarrow 9 + 57 = 6 \quad (\text{False})
  \]

- **For \(X = 4\):**
  \[
  4^2 + 19(4) = 4! \Rightarrow 16 + 76 = 24 \quad (\text{False})
  \]

- **For \(X = 5\):**
  \[
  5^2 + 19(5) = 5! \Rightarrow 25 + 95 = 120 \quad (\text{True})
  \]
  Here, \(120 = 120\), so \(X = 5\) is a solution.

### Conclusion:
The solution to the equation \(X^2 + 19X - X! = 0\) is \(X = 5\).

Would you like more details on the solution, or do you have any questions?

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### 5 Related Questions:
1. What happens if we try non-integer values for \(X\) in this equation?
2. How does the factorial function behave as \(X\) increases?
3. Can there be more than one solution to this type of equation?
4. What methods can be used to solve similar equations that involve factorials?
5. How can we solve quadratic equations when the factorial is not involved?

### Tip:
When dealing with equations that involve factorials, it's often best to test integer values directly, as factorials grow very rapidly, making it easier to find solutions or dismiss non-solutions.

Learn how to solve the equation X^2 + 19X - X! = 0 with a step-by-step solution. This problem involves quadratic equations and factorial functions, requiring careful testing of integer values for X. Suitable for advanced high school students.

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