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The question in the image asks you to solve the quadratic equation \( x^2 + 8x + 5 = 0 \) using the quadratic formula, and select the solutions that apply.

### Quadratic Formula:
The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( x^2 + 8x + 5 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = 8 \)
- \( c = 5 \)

### Step-by-Step Solution:
1. **Calculate the discriminant** \( \Delta \):
   \[
   \Delta = b^2 - 4ac = 8^2 - 4(1)(5) = 64 - 20 = 44
   \]
   
2. **Plug into the quadratic formula**:
   \[
   x = \frac{-8 \pm \sqrt{44}}{2(1)} = \frac{-8 \pm \sqrt{44}}{2}
   \]

3. Simplify the square root of 44:
   \[
   \sqrt{44} = \sqrt{4 \times 11} = 2\sqrt{11}
   \]
   So the solutions become:
   \[
   x = \frac{-8 \pm 2\sqrt{11}}{2}
   \]

4. **Simplify the expression**:
   \[
   x = -4 \pm \sqrt{11}
   \]

### Final Solutions:
The solutions are:
\[
x = -4 + \sqrt{11} \quad \text{or} \quad x = -4 - \sqrt{11}
\]

Looking at the answer choices:
- **C**: \( x = -4 - \sqrt{11} \) is correct.
- **D**: \( x = -4 + \sqrt{11} \) is correct.

The incorrect answers are the ones you selected. You should have selected **C** and **D** instead of **B** and **D**.

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Would you like more details or further clarification on this? Here are some related questions:

1. How do you factor quadratic equations?
2. What are the conditions for a quadratic equation to have real roots?
3. How is the discriminant used to determine the nature of the roots?
4. Can you derive the quadratic formula from completing the square?
5. What are the steps to solve a quadratic equation by factoring?

**Tip**: Always check the sign and value inside the square root when solving using the quadratic formula, to avoid errors.

Solve x^2 + 8x + 5 = 0 using the Quadratic Formula. Select any solutions that apply.

Learn how to solve the quadratic equation x^2 + 8x + 5 using the quadratic formula. This problem demonstrates the step-by-step solution process and covers essential algebraic concepts. Suitable for students in Grades 8-10.

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