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The image you've uploaded contains three quadratic inequalities to solve. Let's address them one by one, detailing the steps:

### 1. Solving \( x^2 + 9x + 14 > 0 \)

**Step 1: Factorize the quadratic expression.**
\[ x^2 + 9x + 14 = (x + 7)(x + 2) \]

**Step 2: Set the expression equal to zero to find critical points.**
\[ (x + 7)(x + 2) = 0 \]
Critical points are \( x = -7 \) and \( x = -2 \).

**Step 3: Test the intervals determined by the critical points.**
The intervals are \( (-\infty, -7) \), \( (-7, -2) \), and \( (-2, \infty) \).

- For \( x < -7 \), choose \( x = -8 \):
  \[ (-8 + 7)(-8 + 2) = (-1)(-6) = 6 > 0 \]
- For \( -7 < x < -2 \), choose \( x = -4 \):
  \[ (-4 + 7)(-4 + 2) = 3 \times (-2) = -6 < 0 \]
- For \( x > -2 \), choose \( x = 0 \):
  \[ (0 + 7)(0 + 2) = 7 \times 2 = 14 > 0 \]

**Solution Interval:**
\[ x \in (-\infty, -7) \cup (-2, \infty) \]

### 2. Solving \( x^2 - 10x + 16 \leq 7 \)

**Step 1: Simplify the inequality.**
\[ x^2 - 10x + 16 - 7 \leq 0 \]
\[ x^2 - 10x + 9 \leq 0 \]

**Step 2: Factorize the quadratic expression.**
\[ x^2 - 10x + 9 = (x - 1)(x - 9) \]

**Step 3: Set the expression equal to zero to find critical points.**
\[ (x - 1)(x - 9) = 0 \]
Critical points are \( x = 1 \) and \( x = 9 \).

**Step 4: Test the intervals determined by the critical points.**
The intervals are \( (-\infty, 1) \), \( (1, 9) \), and \( (9, \infty) \).

- For \( x < 1 \), choose \( x = 0 \):
  \[ (0 - 1)(0 - 9) = (-1)(-9) = 9 > 0 \]
- For \( 1 < x < 9 \), choose \( x = 5 \):
  \[ (5 - 1)(5 - 9) = 4 \times (-4) = -16 < 0 \]
- For \( x > 9 \), choose \( x = 10 \):
  \[ (10 - 1)(10 - 9) = 9 \times 1 = 9 > 0 \]

**Solution Interval:**
\[ x \in [1, 9] \]

### 3. Solving \( x^2 + 6x \leq -5 \)

**Step 1: Simplify the inequality.**
\[ x^2 + 6x + 5 \leq 0 \]

**Step 2: Factorize the quadratic expression.**
\[ x^2 + 6x + 5 = (x + 5)(x + 1) \]

**Step 3: Set the expression equal to zero to find critical points.**
\[ (x + 5)(x + 1) = 0 \]
Critical points are \( x = -5 \) and \( x = -1 \).

**Step 4: Test the intervals determined by the critical points.**
The intervals are \( (-\infty, -5) \), \( (-5, -1) \), and \( (-1, \infty) \).

- For \( x < -5 \), choose \( x = -6 \):
  \[ (-6 + 5)(-6 + 1) = (-1)(-5) = 5 > 0 \]
- For \( -5 < x < -1 \), choose \( x = -3 \):
  \[ (-3 + 5)(-3 + 1) = 2 \times (-2) = -4 < 0 \]
- For \( x > -1 \), choose \( x = 0 \):
  \[ (0 + 5)(0 + 1) = 5 \times 1 = 5 > 0 \]

**Solution Interval:**
\[ x \in [-5, -1] \]

Would you like more details on any of these steps or have any further questions? Here are some related questions you might consider:

1. How can the solutions to these inequalities be graphed on a number line?
2. Why do we test intervals in solving inequalities?
3. What are the implications of using strict and non-strict inequalities in these examples?
4. How would adding or subtracting terms change the solutions?
5. What is the effect of the coefficients of \(x^2\) on the shape of the parabola in these inequalities?

**Tip:** When solving inequalities, always consider using a sign chart to easily visualize the signs of the products in each interval, helping to confirm which intervals satisfy the inequality.

Solve for the solution set of the following quadratic inequalities:
1. x^2 + 9x + 14 > 0
2. x^2 - 10x + 16 < 7
3. x^2 + 6x ≥ -5

Detailed solution for solving quadratic inequalities, including steps for factoring and determining solution intervals. Problems involve inequalities like x^2 + 9x + 14 > 0 and x^2 - 10x + 16 < 7, suitable for students in Grades 9-11.

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