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. **Write the 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 $\frac{1}{2} \times \frac{3x}{x+2} = \frac{3x}{2(x+2)}$. So the equation is: $$x - 3 = \frac{3x}{2(x+2)}$$ 3. **Solve the equation:** Multiply both sides by $2(x+2)$ to clear the denominator: $$2(x+2)(x - 3) = 3x$$ Expand the left side: $$2(x^2 - 3x + 2x - 6) = 3x$$ $$2(x^2 - x - 6) = 3x$$ $$2x^2 - 2x - 12 = 3x$$ Bring all terms to one side: $$2x^2 - 2x - 12 - 3x = 0$$ $$2x^2 - 5x - 12 = 0$$ 4. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=2$, $b=-5$, $c=-12$. Calculate the discriminant: $$\Delta = (-5)^2 - 4(2)(-12) = 25 + 96 = 121$$ Calculate the roots: $$x = \frac{5 \pm \sqrt{121}}{2 \times 2} = \frac{5 \pm 11}{4}$$ So, $$x_1 = \frac{5 + 11}{4} = \frac{16}{4} = 4$$ $$x_2 = \frac{5 - 11}{4} = \frac{-6}{4} = -\frac{3}{2}$$ 5. **Check for restrictions:** The denominator $x+2$ cannot be zero, so $x \neq -2$. Both solutions are valid. 6. **Final answer:** The numbers are $\boxed{4}$ and $\boxed{-\frac{3}{2}}$. This completes the solution using Polya's four-step process.