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The image shows the following equation to solve:

\[
\frac{2x - 3}{x + 4} = 3 - \frac{4x}{x - 2}
\]

### Steps to Solve:

1. **Clear the fractions** by multiplying both sides of the equation by \((x + 4)(x - 2)\), the least common denominator (LCD) of all fractions involved.

   \[
   (x + 4)(x - 2) \cdot \frac{2x - 3}{x + 4} = (x + 4)(x - 2) \cdot \left(3 - \frac{4x}{x - 2}\right)
   \]

2. **Simplify the equation**:

   - The left-hand side simplifies to:
   
     \[
     (x - 2)(2x - 3)
     \]
     
   - The right-hand side expands to:
   
     \[
     3(x + 4)(x - 2) - 4x(x + 4)
     \]

3. **Expand and simplify both sides**:

   - Expanding the left-hand side:
   
     \[
     2x^2 - 4x - 3x + 6 = 2x^2 - 7x + 6
     \]
     
   - Expanding the right-hand side:
   
     \[
     3(x^2 - 2x + 4x - 8) - 4x^2 - 16x
     \]
     Simplifies to:
     
     \[
     3x^2 + 6x - 24 - 4x^2 - 16x = -x^2 - 10x - 24
     \]

4. **Combine like terms** and set the equation to zero:

   \[
   2x^2 - 7x + 6 = -x^2 - 10x - 24
   \]

   Add \(x^2 + 10x + 24\) to both sides:

   \[
   3x^2 + 3x + 30 = 0
   \]

   Simplify by dividing the equation by 3:

   \[
   x^2 + x + 10 = 0
   \]

5. **Solve the quadratic equation** \(x^2 + x + 10 = 0\):

   The discriminant (\(b^2 - 4ac\)) is:

   \[
   1^2 - 4(1)(10) = 1 - 40 = -39
   \]

   Since the discriminant is negative, there are **no real solutions**. The solutions are complex:

   \[
   x = \frac{-1 \pm \sqrt{-39}}{2} = \frac{-1 \pm i\sqrt{39}}{2}
   \]

### Final Answer:
The solutions are complex numbers:
\[
x = \frac{-1 + i\sqrt{39}}{2}, \quad x = \frac{-1 - i\sqrt{39}}{2}
\]

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Learn how to solve the complex equation involving fractions and quadratic expressions step-by-step. This problem covers key algebraic concepts, including equations with fractions, quadratic equations, and complex numbers. Suitable for advanced high school students.

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