1. **State the problem:** Solve the quadratic equation $$\frac{2}{3}x^2 - 5x + \frac{7}{2} = 0$$. 2. **Formula used:** The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. 3. **Identify coefficients:** Here, $a = \frac{2}{3}$, $b = -5$, and $c = \frac{7}{2}$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-5)^2 - 4 \times \frac{2}{3} \times \frac{7}{2} = 25 - \frac{28}{3} = \frac{75}{3} - \frac{28}{3} = \frac{47}{3}$$ 5. **Apply the quadratic formula:** $$x = \frac{-(-5) \pm \sqrt{\frac{47}{3}}}{2 \times \frac{2}{3}} = \frac{5 \pm \sqrt{\frac{47}{3}}}{\frac{4}{3}}$$ 6. **Simplify the denominator by multiplying numerator and denominator by 3:** $$x = \frac{3(5 \pm \sqrt{\frac{47}{3}})}{4} = \frac{15 \pm 3\sqrt{\frac{47}{3}}}{4}$$ 7. **Simplify the square root:** $$3\sqrt{\frac{47}{3}} = 3 \times \frac{\sqrt{47}}{\sqrt{3}} = \frac{3\sqrt{47}}{\sqrt{3}}$$ 8. **Rationalize the denominator:** $$\frac{3\sqrt{47}}{\sqrt{3}} = \frac{3\sqrt{47}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{141}}{3} = \sqrt{141}$$ 9. **Final simplified solutions:** $$x = \frac{15 \pm \sqrt{141}}{4}$$ **Answer:** $$x = \frac{15 + \sqrt{141}}{4} \quad \text{or} \quad x = \frac{15 - \sqrt{141}}{4}$$