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Let's break down the problem from the image and solve it step by step:

### Problem:
You are given the function:

\[
f(x) = (x^2 + 5x + 7)e^x
\]

You need to find the \( x \)-coordinates of the point(s) where the tangent line to the graph of \( f(x) \) is horizontal. This means you need to find where the derivative of the function \( f'(x) \) is equal to zero.

### Steps to solve:

1. **Differentiate \( f(x) \)**:
   - The given function is a product of two functions: \( (x^2 + 5x + 7) \) and \( e^x \). Use the product rule to differentiate.

   The product rule states:

   \[
   \frac{d}{dx}(u \cdot v) = u'v + uv'
   \]

   Here, \( u = x^2 + 5x + 7 \) and \( v = e^x \).

   - First, differentiate \( u \) and \( v \):
     \[
     u' = 2x + 5
     \]
     \[
     v' = e^x
     \]

   - Now apply the product rule:
     \[
     f'(x) = (2x + 5)e^x + (x^2 + 5x + 7)e^x
     \]

   This simplifies to:

   \[
   f'(x) = e^x \left( (2x + 5) + (x^2 + 5x + 7) \right)
   \]

   Combine the terms inside the parentheses:

   \[
   f'(x) = e^x \left( x^2 + 7x + 12 \right)
   \]

2. **Set the derivative equal to zero**:
   For the tangent to be horizontal, the derivative must be zero:

   \[
   e^x(x^2 + 7x + 12) = 0
   \]

   Since \( e^x \) is never zero, we only need to solve:

   \[
   x^2 + 7x + 12 = 0
   \]

3. **Solve the quadratic equation**:
   Solve \( x^2 + 7x + 12 = 0 \) using the quadratic formula:

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

   Here, \( a = 1 \), \( b = 7 \), and \( c = 12 \). Plugging in these values:

   \[
   x = \frac{-7 \pm \sqrt{7^2 - 4(1)(12)}}{2(1)}
   \]
   \[
   x = \frac{-7 \pm \sqrt{49 - 48}}{2}
   \]
   \[
   x = \frac{-7 \pm \sqrt{1}}{2}
   \]
   \[
   x = \frac{-7 \pm 1}{2}
   \]

   This gives two solutions:

   \[
   x = \frac{-7 + 1}{2} = \frac{-6}{2} = -3
   \]
   \[
   x = \frac{-7 - 1}{2} = \frac{-8}{2} = -4
   \]

### Final Answer:
The \( x \)-coordinates where the tangent is horizontal are \( x = -3 \) and \( x = -4 \). Therefore, the answer is:

\[
x = -4; -3
\]

Would you like more details on any step or have other questions?

### Related Questions:
1. How does the product rule work when differentiating two functions?
2. Why is the term \( e^x \) never zero, and how does that affect the equation?
3. Can you explain more about solving quadratic equations using the quadratic formula?
4. How do horizontal tangents relate to derivative concepts in calculus?
5. What are some real-world applications of finding horizontal tangents in graphs?

**Tip:** When solving a product rule differentiation problem, ensure you differentiate both parts of the product separately before combining the terms.

Find the x-coordinate(s) of the point(s) where the tangent line to the graph of f(x) = (x^2 + 5x + 7) e^x is horizontal.

This problem involves finding the x-coordinate(s) where the tangent to the graph of f(x) = (x^2 + 5x + 7)e^x is horizontal. It requires knowledge of the product rule, differentiation, and solving quadratic equations. Suitable for college-level calculus students.

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