1. **State the problem:** Find all numbers $x$ such that the difference of the number and 3 equals half of the quotient of thrice the number and the sum of the number and two. 2. **Devise a plan:** Translate the problem into an equation. The difference of a number and 3 is $x - 3$. The quotient of thrice the number and the sum of the number and two is $\frac{3x}{x+2}$. Half of this quotient is $\frac{1}{2} \times \frac{3x}{x+2} = \frac{3x}{2(x+2)}$. Set these equal: $x - 3 = \frac{3x}{2(x+2)}$. 3. **Carry out the plan:** Multiply both sides by $2(x+2)$ to clear the denominator: $$2(x+2)(x - 3) = 3x$$ Expand the left side: $$2(x^2 + 2x - 3x - 6) = 3x$$ Simplify inside parentheses: $$2(x^2 - x - 6) = 3x$$ Distribute 2: $$2x^2 - 2x - 12 = 3x$$ Bring all terms to one side: $$2x^2 - 2x - 12 - 3x = 0$$ $$2x^2 - 5x - 12 = 0$$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=2$, $b=-5$, $c=-12$: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(-12)}}{2 \times 2} = \frac{5 \pm \sqrt{25 + 96}}{4} = \frac{5 \pm \sqrt{121}}{4}$$ $$x = \frac{5 \pm 11}{4}$$ Two solutions: $$x = \frac{5 + 11}{4} = \frac{16}{4} = 4$$ $$x = \frac{5 - 11}{4} = \frac{-6}{4} = -\frac{3}{2}$$ 4. **Look back and reflect:** Check solutions in the original equation: For $x=4$: Left side: $4 - 3 = 1$ Right side: $\frac{3 \times 4}{2(4+2)} = \frac{12}{2 \times 6} = \frac{12}{12} = 1$ For $x = -\frac{3}{2}$: Left side: $-\frac{3}{2} - 3 = -\frac{3}{2} - \frac{6}{2} = -\frac{9}{2}$ Right side: $\frac{3 \times (-\frac{3}{2})}{2(-\frac{3}{2} + 2)} = \frac{-\frac{9}{2}}{2(\frac{1}{2})} = \frac{-\frac{9}{2}}{1} = -\frac{9}{2}$ Both satisfy the equation. **Final answer:** The numbers are $\boxed{4}$ and $\boxed{-\frac{3}{2}}$.