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The inequality provided is:

\[
(x + 3)(x - 7) > 0
\]

To solve this inequality, follow these steps:

### Step 1: Find the critical points
The expression will equal zero at the values of \(x\) that make either factor zero:
- \(x + 3 = 0 \Rightarrow x = -3\)
- \(x - 7 = 0 \Rightarrow x = 7\)

So, the critical points are \(x = -3\) and \(x = 7\).

### Step 2: Test the intervals
These two points divide the number line into three intervals:
1. \( (-\infty, -3) \)
2. \( (-3, 7) \)
3. \( (7, \infty) \)

Now, we need to test points in each of these intervals to see where the inequality holds:

- For \(x < -3\), choose \(x = -4\):  
  \((x + 3)(x - 7) = (-4 + 3)(-4 - 7) = (-1)(-11) = 11 > 0\), so the inequality is true in this interval.
  
- For \(-3 < x < 7\), choose \(x = 0\):  
  \((x + 3)(x - 7) = (0 + 3)(0 - 7) = (3)(-7) = -21 < 0\), so the inequality is false in this interval.
  
- For \(x > 7\), choose \(x = 8\):  
  \((x + 3)(x - 7) = (8 + 3)(8 - 7) = (11)(1) = 11 > 0\), so the inequality is true in this interval.

### Step 3: Write the solution
The solution is the union of the intervals where the inequality is true:
\[
x \in (-\infty, -3) \cup (7, \infty)
\]

### Step 4: Graph the solution
- Draw an open circle at \(x = -3\) and \(x = 7\) to indicate these points are not included in the solution.
- Shade the region to the left of \(-3\) and to the right of \(7\).

Would you like any further details or help on this?

### Follow-up Questions:
1. How would the solution change if the inequality were \( \geq \) instead of \( > \)?
2. What happens if the inequality had been \( (x + 3)(x - 7) < 0 \)?
3. How do you find the critical points for more complicated inequalities?
4. Why do we use open circles at \(x = -3\) and \(x = 7\) for this inequality?
5. Can we solve this inequality using a different method, such as a sign chart?

### Tip:
Always test points in each interval when solving inequalities involving quadratic expressions to determine where the inequality holds.

Learn how to solve and graph the quadratic inequality (x + 3)(x - 7) > 0 by finding critical points and testing intervals. This problem covers important algebra concepts and is suitable for Grades 9-12.

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