To solve the equation $x^2 + 5x + 8 = 4$ by completing the square, we first rearrange it:

1. Move 4 to the left side: $x^2 + 5x + 8 - 4 = 0 \implies x^2 + 5x + 4 = 0$

2. Next, we complete the square. We focus on the quadratic and linear terms: $x^2 + 5x$ To complete the square, take half of the coefficient of $x$ (which is $5$), square it, and add it to both sides. Half of $5$ is $\frac{5}{2}$, and squaring it gives: $\left(\frac{5}{2}\right)^2 = \frac{25}{4}$

3. Add $\frac{25}{4}$ to both sides of the equation: $x^2 + 5x + \frac{25}{4} = 4 + \frac{25}{4}$

4. Simplify the right side: $4 = \frac{16}{4} \implies 4 + \frac{25}{4} = \frac{16}{4} + \frac{25}{4} = \frac{41}{4}$

5. Now, the equation is: $x^2 + 5x + \frac{25}{4} = \frac{41}{4}$

6. The left side can be factored as a perfect square: $\left(x + \frac{5}{2}\right)^2 = \frac{41}{4}$

7. Taking the square root of both sides gives: $x + \frac{5}{2} = \pm \sqrt{\frac{41}{4}} \implies x + \frac{5}{2} = \pm \frac{\sqrt{41}}{2}$

8. Finally, solving for $x$: $x = -\frac{5}{2} \pm \frac{\sqrt{41}}{2}$

So, the solutions are: $x = -\frac{5 + \sqrt{41}}{2} \quad \text{and} \quad x = -\frac{5 - \sqrt{41}}{2}$

Would you like further details or clarification on any part of this process? Here are some related questions:

1. What does completing the square mean?
2. How can we derive the quadratic formula from this method?
3. Can you show another example of completing the square?
4. How does the discriminant relate to the solutions of a quadratic equation?
5. What are the geometric interpretations of quadratic equations?

Tip: When completing the square, always remember to add and subtract the same value to keep the equation balanced.