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The problem shown in the image asks to find the roots of the given quadratic equation. The equation is:

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

### Step-by-step Solution:

1. **Expand the right-hand side**:
   \[
   \frac{4}{3} \left( 2 - \frac{x}{2} \right) = \frac{4}{3} \cdot 2 - \frac{4}{3} \cdot \frac{x}{2} = \frac{8}{3} - \frac{4x}{6} = \frac{8}{3} - \frac{2x}{3}
   \]
   So, the equation becomes:
   \[
   x^2 - 4x = \frac{8}{3} - \frac{2x}{3}
   \]

2. **Multiply through by 3 to eliminate the fractions**:
   \[
   3(x^2 - 4x) = 3\left(\frac{8}{3} - \frac{2x}{3}\right)
   \]
   This simplifies to:
   \[
   3x^2 - 12x = 8 - 2x
   \]

3. **Move all terms to one side of the equation**:
   \[
   3x^2 - 12x + 2x - 8 = 0
   \]
   Simplifying:
   \[
   3x^2 - 10x - 8 = 0
   \]

4. **Solve the quadratic equation** using the quadratic formula:
   \[
   x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(-8)}}{2(3)}
   \]
   \[
   x = \frac{10 \pm \sqrt{100 + 96}}{6}
   \]
   \[
   x = \frac{10 \pm \sqrt{196}}{6}
   \]
   \[
   x = \frac{10 \pm 14}{6}
   \]
   Therefore, the two solutions are:
   \[
   x = \frac{10 + 14}{6} = \frac{24}{6} = 4
   \]
   \[
   x = \frac{10 - 14}{6} = \frac{-4}{6} = \frac{-2}{3}
   \]

### Answer:
The roots are \( x = 4 \) and \( x = \frac{-2}{3} \), which corresponds to option **C**.

---

Would you like further details on any step? Here are some related questions to practice:

1. How do you apply the quadratic formula to equations?
2. What are the benefits of eliminating fractions in quadratic equations?
3. How does multiplying both sides of an equation help in solving?
4. Can the quadratic formula solve all quadratic equations?
5. How do you verify the roots of a quadratic equation?

**Tip**: Always check your solutions by substituting the roots back into the original equation.

Find the roots of the quadratic equation: x^2 - 4x = (4/3)(2 - x/2)

Learn how to solve the quadratic equation x^2 - 4x = (4/3)(2 - x/2) with a detailed step-by-step solution. This problem covers key algebraic concepts such as expanding expressions, eliminating fractions, and using the quadratic formula. Suitable for students in Grades 9-10.

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