https://file.aiquickdraw.com/mathbot/file/1725840383101i4vvrpvt.jpg

The equation given in the image is:

\[
(x + 6)^2 + 12(x + 6) + 35 = 0
\]

We are tasked with making an appropriate substitution and rewriting the equation in quadratic form.

### Step 1: Substitution
Let:
\[
u = x + 6
\]

This simplifies the equation to:

\[
u^2 + 12u + 35 = 0
\]

### Step 2: Solve the quadratic equation
Now, solve the quadratic equation:
\[
u^2 + 12u + 35 = 0
\]

We can use factoring to solve it:

\[
(u + 7)(u + 5) = 0
\]

So, the solutions for \( u \) are:
\[
u = -7 \quad \text{or} \quad u = -5
\]

### Step 3: Back-substitute \( u = x + 6 \)
Since \( u = x + 6 \), we substitute back to find \( x \):

1. For \( u = -7 \):
   \[
   x + 6 = -7 \quad \Rightarrow \quad x = -13
   \]

2. For \( u = -5 \):
   \[
   x + 6 = -5 \quad \Rightarrow \quad x = -11
   \]

### Final Answer:
The solutions for \( x \) are:
\[
x = -13 \quad \text{or} \quad x = -11
\]

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

### 5 Related Questions:
1. How do you factor a quadratic equation like \( u^2 + 12u + 35 = 0 \)?
2. Can you explain the process of substitution in more detail?
3. What if the quadratic equation cannot be factored easily? How would you solve it?
4. How does completing the square work for quadratic equations?
5. Can you use the quadratic formula to solve the equation as well?

### Tip:
When factoring quadratic equations, always check if the middle term can be split into factors of the constant term that add up to the coefficient of the linear term.

Solve the following equation by making an appropriate substitution: (x + 6)^2 + 12(x + 6) + 35 = 0

Learn how to solve the quadratic equation (x + 6)^2 + 12(x + 6) + 35 = 0 using substitution. This problem covers key algebra concepts including quadratic equations and factoring methods, suitable for students in Grades 8-10.

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