We are tasked with solving the equation:

\[
\left( \frac{x}{x - 2} \right)^2 - 2 \left( \frac{x}{x - 2} \right) - 15 = 0
\]

### Step 1: Substitution
Let \( y = \frac{x}{x - 2} \). This simplifies the equation to a quadratic form in \( y \):

\[
y^2 - 2y - 15 = 0
\]

### Step 2: Solve the quadratic equation
Now, we solve the quadratic equation:

\[
y^2 - 2y - 15 = 0
\]

We can factor this:

\[
(y - 5)(y + 3) = 0
\]

Thus, \( y = 5 \) or \( y = -3 \).

### Step 3: Back-substitute \( y = \frac{x}{x - 2} \)
We now return to the original variable \( x \). Recall that \( y = \frac{x}{x - 2} \), so we solve for \( x \) in both cases.

#### Case 1: \( y = 5 \)
\[
\frac{x}{x - 2} = 5
\]

Multiply both sides by \( x - 2 \):

\[
x = 5(x - 2)
\]

Simplify:

\[
x = 5x - 10
\]

\[
-4x = -10
\]

\[
x = \frac{10}{4} = \frac{5}{2}
\]

#### Case 2: \( y = -3 \)
\[
\frac{x}{x - 2} = -3
\]

Multiply both sides by \( x - 2 \):

\[
x = -3(x - 2)
\]

Simplify:

\[
x = -3x + 6
\]

\[
4x = 6
\]

\[
x = \frac{6}{4} = \frac{3}{2}
\]

### Final Answer:
The solutions to the equation are:

\[
x = \frac{5}{2} \quad \text{or} \quad x = \frac{3}{2}
\]

Would you like further details on any of these steps?

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Here are 5 related questions for you to explore:
1. How can you derive the quadratic formula from a general quadratic equation?
2. What happens when the denominator in \( \frac{x}{x-2} \) is zero?
3. Can we solve this type of equation without substitution?
4. How would the solution change if the equation had complex numbers?
5. How can we verify these solutions by plugging them back into the original equation?

**Tip:** Always check your solutions by substituting them back into the original equation to ensure no division by zero or incorrect results occur.

This detailed guide explains how to solve the equation (x/(x-2))^2 - 2(x/(x-2)) - 15 = 0 using substitution and factoring methods. Learn how to apply the quadratic formula and substitution to simplify complex algebraic expressions. Suitable for students in Grades 9-11.

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