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The problem provided involves a quadratic equation \(x^2 + 20x = 69\). We need to determine which of the given values of \(x\) (i.e., \(x = -23\), \(x = 0\), and \(x = 12\)) satisfies the equation by substituting them into the equation, evaluating the result, and checking if the equation holds true.

Let's substitute each value into the equation and check the result:

1. **For \(x = -23\):**
   - Substituting into \(x^2 + 20x = 69\):
   \[
   (-23)^2 + 20(-23) = 529 - 460 = 69
   \]
   - The equation holds true for \(x = -23\).

2. **For \(x = 0\):**
   - Substituting into \(x^2 + 20x = 69\):
   \[
   0^2 + 20(0) = 0
   \]
   - The result is 0, which is not equal to 69, so \(x = 0\) is not a solution.

3. **For \(x = 12\):**
   - Substituting into \(x^2 + 20x = 69\):
   \[
   12^2 + 20(12) = 144 + 240 = 384
   \]
   - The result is 384, which is not equal to 69, so \(x = 12\) is not a solution.

### Conclusion:
Only \(x = -23\) is in the solution set.

Would you like more details or have any questions?

Here are 5 related questions:
1. How can we solve quadratic equations using factoring?
2. What is the quadratic formula, and how does it apply to this problem?
3. How would the solution change if the equation were \(x^2 + 20x = -69\)?
4. What happens when a quadratic equation has no real solutions?
5. Can a quadratic equation have more than two real solutions?

**Tip:** Always check both possible solutions when solving a quadratic equation; some equations may have two valid solutions.

For the quadratic equation x^2 + 20x = 69, complete the table by testing the given values to determine which one is in the solution set.

Learn how to solve the quadratic equation x^2 + 20x = 69 by substituting different values of x to check for solutions. This algebra problem is ideal for students in Grades 7-9, helping them understand quadratic equations and their solutions.

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