To solve the equation \((4x + 7)^2 = 64\), we follow these steps:

1. **Take the square root of both sides**:  
   \[
   \sqrt{(4x + 7)^2} = \pm\sqrt{64}
   \]
   This simplifies to:
   \[
   4x + 7 = \pm 8
   \]

2. **Set up two separate equations** based on the positive and negative square root values:

   1. \( 4x + 7 = 8 \)
   2. \( 4x + 7 = -8 \)

3. **Solve each equation for \( x \):**

   For \( 4x + 7 = 8 \):
   \[
   4x = 8 - 7 \implies 4x = 1 \implies x = \frac{1}{4}
   \]

   For \( 4x + 7 = -8 \):
   \[
   4x = -8 - 7 \implies 4x = -15 \implies x = -\frac{15}{4}
   \]

4. **Solution set**:  
   The solution set for the equation is:
   \[
   \left\{ \frac{1}{4}, -\frac{15}{4} \right\}
   \]

Would you like more details on any of these steps, or do you have any other questions?

Here are some related questions:

1. How do you solve similar equations with different coefficients?
2. What is the geometric interpretation of squaring an expression?
3. How can we verify solutions by substituting back into the original equation?
4. What happens if the squared term is negative?
5. How would this equation change if we had \((4x + 7)^2 \leq 64\)?

**Tip:** Always remember to consider both positive and negative square roots when solving equations like \((ax + b)^2 = c\).

Learn how to solve the equation (4x + 7)^2 = 64 with a detailed step-by-step solution. This algebraic problem covers solving equations using square roots and basic linear equations. Suitable for students in Grades 8-10.

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