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The image shows a set of quadratic inequalities to be solved. The first few inequalities are solved as examples in the table, and there are additional exercises below. Let me guide you through solving these inequalities one by one:

1. **\( x^2 \leq 0 \)**  
   The inequality \( x^2 \leq 0 \) is satisfied only when \( x = 0 \), since \( x^2 \) is always non-negative and equal to 0 only at \( x = 0 \).  
   **Solution**: \( x = 0 \).

2. **\( -x^2 \geq 0 \)**  
   This is equivalent to \( x^2 \leq 0 \), which, like the previous one, is only true when \( x = 0 \).  
   **Solution**: \( x = 0 \).

3. **\( x^2 + 3x - 18 \leq 0 \)**  
   First, solve the quadratic equation \( x^2 + 3x - 18 = 0 \) using the quadratic formula:  
   \[
   x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-18)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 72}}{2} = \frac{-3 \pm \sqrt{81}}{2} = \frac{-3 \pm 9}{2}
   \]
   So, \( x_1 = 3 \) and \( x_2 = -6 \). Now, we analyze the intervals:  
   - For \( x < -6 \), \( x^2 + 3x - 18 > 0 \).  
   - For \( -6 < x < 3 \), \( x^2 + 3x - 18 \leq 0 \).  
   - For \( x > 3 \), \( x^2 + 3x - 18 > 0 \).  
   **Solution**: \( -6 \leq x \leq 3 \).

4. **\( 2x^2 + 7x < 0 \)**  
   Solve the quadratic equation \( 2x^2 + 7x = 0 \) by factoring:  
   \[
   x(2x + 7) = 0
   \]
   So, \( x = 0 \) and \( x = -\frac{7}{2} \). Now, analyze the intervals:  
   - For \( x < -\frac{7}{2} \), \( 2x^2 + 7x > 0 \).  
   - For \( -\frac{7}{2} < x < 0 \), \( 2x^2 + 7x < 0 \).  
   - For \( x > 0 \), \( 2x^2 + 7x > 0 \).  
   **Solution**: \( -\frac{7}{2} < x < 0 \).

5. **\( x^2 + 16 \leq 0 \)**  
   Since \( x^2 \) is always non-negative and \( x^2 + 16 \) is strictly greater than 0 for all real values of \( x \), there is no real solution to this inequality.  
   **Solution**: No solution.

6. **\( -3x^2 - 5x - 2 \leq 0 \)**  
   Solve the quadratic equation \( -3x^2 - 5x - 2 = 0 \) using the quadratic formula:  
   \[
   x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(-3)(-2)}}{2(-3)} = \frac{5 \pm \sqrt{25 - 24}}{-6} = \frac{5 \pm 1}{-6}
   \]
   So, \( x_1 = -1 \) and \( x_2 = -\frac{2}{3} \). Now, analyze the intervals:  
   - For \( x < -1 \), \( -3x^2 - 5x - 2 \leq 0 \).  
   - For \( -1 \leq x \leq -\frac{2}{3} \), \( -3x^2 - 5x - 2 \leq 0 \).  
   - For \( x > -\frac{2}{3} \), \( -3x^2 - 5x - 2 > 0 \).  
   **Solution**: \( -1 \leq x \leq -\frac{2}{3} \).

7. **\( 3x^2 + 12 \leq 0 \)**  
   Like in problem 5, since \( 3x^2 + 12 \) is always positive, there is no real solution.  
   **Solution**: No solution.

Would you like any specific details or clarifications for these solutions? Here are some related questions:

1. How do you find the solutions for quadratic inequalities with two real roots?
2. Can you explain the graphical interpretation of quadratic inequalities?
3. What is the difference between strict and non-strict inequalities in quadratic equations?
4. Why do some quadratic inequalities have no solution?
5. How can you solve inequalities using the quadratic formula?

**Tip**: Always plot the quadratic function to better understand the regions where the inequality holds.

Solve the following quadratic inequalities presented in the image:

Learn how to solve quadratic inequalities such as x² ≤ 0, -x² ≥ 0, and more. Detailed solutions include factoring and using the quadratic formula, with graphical interpretations of the inequalities. Suitable for students in Grades 9-12.

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