We can solve the equation \((x + 3)^2 + 11(x + 3) - 12 = 0\) by using substitution. Let's define:

\[
y = x + 3
\]

So the equation becomes:

\[
y^2 + 11y - 12 = 0
\]

This is now a quadratic equation in terms of \(y\). We can solve it using the quadratic formula:

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

For the equation \(y^2 + 11y - 12 = 0\), we have:

- \(a = 1\)
- \(b = 11\)
- \(c = -12\)

Now, substitute these values into the quadratic formula:

\[
y = \frac{-11 \pm \sqrt{11^2 - 4(1)(-12)}}{2(1)}
\]
\[
y = \frac{-11 \pm \sqrt{121 + 48}}{2}
\]
\[
y = \frac{-11 \pm \sqrt{169}}{2}
\]
\[
y = \frac{-11 \pm 13}{2}
\]

So we have two possible solutions for \(y\):

\[
y = \frac{-11 + 13}{2} = \frac{2}{2} = 1
\]
\[
y = \frac{-11 - 13}{2} = \frac{-24}{2} = -12
\]

Now, recall that \(y = x + 3\), so we can solve for \(x\):

1. If \(y = 1\), then:
\[
x + 3 = 1 \quad \Rightarrow \quad x = 1 - 3 = -2
\]

2. If \(y = -12\), then:
\[
x + 3 = -12 \quad \Rightarrow \quad x = -12 - 3 = -15
\]

Thus, the two solutions are:

\[
x = -2 \quad \text{or} \quad x = -15
\]

### Final answer:
\[
x = -2 \quad \text{or} \quad x = -15
\]

Would you like more details on this solution, or do you have any questions?

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Here are 5 follow-up questions related to this:

1. How do you apply the quadratic formula to different forms of equations?
2. What is the process for completing the square to solve quadratic equations?
3. Can you describe how substitution helps simplify complex equations?
4. How can you check whether your solutions for \(x\) are correct?
5. What is the discriminant and how does it determine the nature of the roots?

**Tip:** Always simplify your equation first before deciding which method (like substitution or factoring) to use for solving it.

Learn how to solve the quadratic equation (x+3)^2 + 11(x+3) - 12 = 0 using the substitution method and the quadratic formula. This step-by-step solution covers key algebraic concepts such as substitution and applying the quadratic formula. Suitable for students in Grades 8-10.

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