1. **State the problem:** Solve the inequality $$x^2 - 3x > \frac{1}{2}x^2 - 4$$. 2. **Rewrite the inequality:** Move all terms to one side to compare with zero. $$x^2 - 3x - \frac{1}{2}x^2 + 4 > 0$$ 3. **Combine like terms:** $$\left(x^2 - \frac{1}{2}x^2\right) - 3x + 4 > 0$$ $$\frac{1}{2}x^2 - 3x + 4 > 0$$ 4. **Multiply both sides by 2 to clear the fraction:** $$2 \times \left(\frac{1}{2}x^2 - 3x + 4\right) > 2 \times 0$$ $$\cancel{2} \times \frac{1}{\cancel{2}} x^2 - 6x + 8 > 0$$ $$x^2 - 6x + 8 > 0$$ 5. **Factor the quadratic:** $$x^2 - 6x + 8 = (x - 2)(x - 4)$$ 6. **Analyze the inequality:** We want to find where $$(x - 2)(x - 4) > 0$$ 7. **Determine intervals:** - The roots are $x=2$ and $x=4$. - The quadratic opens upwards (coefficient of $x^2$ is positive). 8. **Sign analysis:** - For $x < 2$, both $(x - 2)$ and $(x - 4)$ are negative, so their product is positive. - For $2 < x < 4$, one factor is positive and the other negative, so product is negative. - For $x > 4$, both factors are positive, so product is positive. 9. **Solution:** $$x < 2 \quad \text{or} \quad x > 4$$ **Final answer:** $$\boxed{x < 2 \text{ or } x > 4}$$