Let's solve the given expression:

\[
(y)^{2} - 11y + 24
\]

### Step 1: Substitute the expression \( y = x^2 - 2x \) into the equation.

The equation becomes:

\[
(y)^{2} - 11y + 24 = 0
\]

### Step 2: Factor the quadratic equation.

We need to factorize the quadratic equation \( y^2 - 11y + 24 = 0 \). To do this, we find two numbers that multiply to 24 (the constant term) and add to -11 (the coefficient of \( y \)).

The numbers -8 and -3 satisfy these conditions because:

\[
-8 \times -3 = 24 \quad \text{and} \quad -8 + -3 = -11
\]

So, we can factorize the equation as:

\[
(y - 8)(y - 3) = 0
\]

### Step 3: Solve for \( y \).

Set each factor equal to zero:

\[
y - 8 = 0 \quad \text{or} \quad y - 3 = 0
\]

So:

\[
y = 8 \quad \text{or} \quad y = 3
\]

### Step 4: Substitute back \( y = x^2 - 2x \).

We now solve for \( x \) in the two cases:

1. **Case 1:** \( x^2 - 2x = 8 \)

   \[
   x^2 - 2x - 8 = 0
   \]

   Factorizing this:

   \[
   (x - 4)(x + 2) = 0
   \]

   Therefore, \( x = 4 \) or \( x = -2 \).

2. **Case 2:** \( x^2 - 2x = 3 \)

   \[
   x^2 - 2x - 3 = 0
   \]

   Factorizing this:

   \[
   (x - 3)(x + 1) = 0
   \]

   Therefore, \( x = 3 \) or \( x = -1 \).

### Final Solutions:

The solutions for \( x \) are:

\[
x = 4, \quad x = -2, \quad x = 3, \quad \text{and} \quad x = -1
\]

Would you like to explore any details or have any questions?

Here are 5 related questions:

1. How do you factorize a quadratic equation?
2. What is the quadratic formula and when do you use it?
3. How can you check the solutions of a quadratic equation?
4. What are the steps to solve higher-degree polynomial equations?
5. How does completing the square work in solving quadratic equations?

**Tip:** Always verify your solutions by substituting them back into the original equation to ensure accuracy.

Learn how to solve the quadratic equation (x^2 - 2x)^2 - 11(x^2 - 2x) + 24 with a detailed step-by-step solution. This problem covers key algebraic concepts, including factoring quadratics and substitution. Suitable for students in Grades 7-9.

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