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We are tasked with solving for \( x \) in the given equation:

\[
4(7 + 5x^2) = 348
\]

### Step 1: Simplify the equation
First, distribute the 4 across the terms inside the parentheses:

\[
4 \times 7 + 4 \times 5x^2 = 348
\]
\[
28 + 20x^2 = 348
\]

### Step 2: Isolate the \( x^2 \) term
Subtract 28 from both sides to simplify further:

\[
20x^2 = 348 - 28
\]
\[
20x^2 = 320
\]

### Step 3: Solve for \( x^2 \)
Now, divide both sides by 20:

\[
x^2 = \frac{320}{20}
\]
\[
x^2 = 16
\]

### Step 4: Solve for \( x \)
Take the square root of both sides:

\[
x = \pm \sqrt{16}
\]
\[
x = \pm 4
\]

### Final Answer:
The two possible values of \( x \) are:

\[
x = 4 \quad \text{or} \quad x = -4
\]

Let me know if you would like more details or have any questions!

Here are 5 related questions to deepen your understanding:
1. How do we check if both solutions \( x = 4 \) and \( x = -4 \) satisfy the original equation?
2. What would happen if we had a negative constant on the right side of the equation?
3. How does the equation change if we modify the coefficient of \( x^2 \)?
4. What are some real-world applications of solving quadratic equations?
5. How can the quadratic formula be used to solve this problem?

**Tip:** When solving quadratic equations, always consider both positive and negative square root values unless the problem specifies otherwise.

Calculate the two possible values of x in the following equation: 4(7 + 5x^2) = 348

Discover how to solve the quadratic equation 4(7 + 5x^2) = 348 step-by-step. This example demonstrates key algebraic concepts, including expanding, isolating terms, and taking square roots. Appropriate for Grades 9-11.

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