Solve the equation: $$\frac{3(x-6)}{2} - \frac{x^2}{3} = \frac{3}{1}$$.

Let's solve the equation step by step: 1. Multiply both sides of the equation by 6 to eliminate the fractions: $$6 \cdot \left( \frac{3(x-6)}{2} - \frac{x^2}{3} \right) = 6 \cdot 3$$ $$9(x-6) - 2x^2 = 18$$ 2. Expand the expression: $$9x - 54 - 2x^2 = 18$$ 3. Rearrange the equation to form a quadratic equation: $$-2x^2 + 9x - 54 - 18 = 0$$ $$-2x^2 + 9x - 72 = 0$$ 4. Multiply the equation by -1 to make the leading coefficient positive: $$2x^2 - 9x + 72 = 0$$ 5. Use the quadratic formula to solve for x: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where a = 2, b = -9, and c = 72. 6. Calculate the discriminant (\(D\)): $$D = b^2 - 4ac$$ $$D = (-9)^2 - 4(2)(72)$$ $$D = 81 - 576$$ $$D = -495$$ 7. Since the discriminant is negative (\(D < 0\)), there are no real solutions for the quadratic equation. Therefore, the equation has no real solutions. Answer: No real solutions