To solve the equation:

\[
\frac{x^2}{x - 3} = \frac{x + 2}{2x - 5}
\]

we will use cross-multiplication to eliminate the denominators.

### Step 1: Cross-multiply
\[
(x^2)(2x - 5) = (x - 3)(x + 2)
\]

### Step 2: Expand both sides

#### Left-hand side:
\[
x^2(2x - 5) = x^2 \cdot 2x - x^2 \cdot 5 = 2x^3 - 5x^2
\]

#### Right-hand side:
\[
(x - 3)(x + 2) = x^2 + 2x - 3x - 6 = x^2 - x - 6
\]

Now, the equation becomes:
\[
2x^3 - 5x^2 = x^2 - x - 6
\]

### Step 3: Move all terms to one side
\[
2x^3 - 5x^2 - x^2 + x + 6 = 0
\]
Simplifying:
\[
2x^3 - 6x^2 + x + 6 = 0
\]

### Step 4: Solve the cubic equation

We now have a cubic equation:
\[
2x^3 - 6x^2 + x + 6 = 0
\]
We will try to find rational roots using the Rational Root Theorem. Testing possible values like \( x = 2 \), \( x = -1 \), etc.

Let me calculate that for you.

The solutions to the cubic equation are:

\[
x = 2, \quad x = \frac{1}{2} - \frac{\sqrt{7}}{2}, \quad x = \frac{1}{2} + \frac{\sqrt{7}}{2}
\]

Thus, the three solutions for \(x\) are \(2\), and the two irrational numbers \( \frac{1}{2} - \frac{\sqrt{7}}{2} \) and \( \frac{1}{2} + \frac{\sqrt{7}}{2} \).

Do you have any questions or would you like further details?

### Related Questions:
1. How do you apply the Rational Root Theorem?
2. How can you check if the solutions are extraneous?
3. What is the process to solve general cubic equations?
4. How do irrational numbers affect the solution?
5. How can graphing help visualize the solution to cubic equations?

**Tip:** Always check for extraneous solutions when solving equations that involve fractions or roots.

This problem involves solving the cubic equation x²/(x-3) = (x+2)/(2x-5) using cross-multiplication and the Rational Root Theorem. Follow the step-by-step process to expand, simplify, and find the solutions, suitable for students in Grades 10-12.

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