1. **State the problem:** We have two resistors in parallel with total resistance $R_T = 0.75$ ohms. One resistor $R_2$ is 2 ohms more than the other resistor $R_1$, so $R_2 = R_1 + 2$. We want to find $R_1$ and $R_2$ given the formula for total resistance in parallel: $$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}$$ 2. **Write the formula with given values:** $$\frac{1}{0.75} = \frac{1}{R_1} + \frac{1}{R_1 + 2}$$ 3. **Simplify the left side:** $$\frac{1}{0.75} = \frac{4}{3}$$ So, $$\frac{4}{3} = \frac{1}{R_1} + \frac{1}{R_1 + 2}$$ 4. **Find common denominator and combine right side:** $$\frac{1}{R_1} + \frac{1}{R_1 + 2} = \frac{(R_1 + 2) + R_1}{R_1(R_1 + 2)} = \frac{2R_1 + 2}{R_1(R_1 + 2)}$$ 5. **Set equation:** $$\frac{4}{3} = \frac{2R_1 + 2}{R_1(R_1 + 2)}$$ 6. **Cross multiply:** $$4 \cdot R_1 (R_1 + 2) = 3 (2R_1 + 2)$$ 7. **Expand both sides:** $$4 (R_1^2 + 2R_1) = 6R_1 + 6$$ $$4R_1^2 + 8R_1 = 6R_1 + 6$$ 8. **Bring all terms to one side:** $$4R_1^2 + 8R_1 - 6R_1 - 6 = 0$$ $$4R_1^2 + 2R_1 - 6 = 0$$ 9. **Simplify by dividing entire equation by 2:** $$\cancel{2} (2R_1^2 + R_1 - 3) = 0$$ $$2R_1^2 + R_1 - 3 = 0$$ 10. **Solve quadratic equation:** Use quadratic formula: $$R_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=1$, $c=-3$. Calculate discriminant: $$\Delta = 1^2 - 4 \times 2 \times (-3) = 1 + 24 = 25$$ Calculate roots: $$R_1 = \frac{-1 \pm \sqrt{25}}{2 \times 2} = \frac{-1 \pm 5}{4}$$ 11. **Find possible values:** $$R_1 = \frac{-1 + 5}{4} = \frac{4}{4} = 1$$ $$R_1 = \frac{-1 - 5}{4} = \frac{-6}{4} = -1.5$$ Since resistance must be positive, discard $R_1 = -1.5$. 12. **Find $R_2$:** $$R_2 = R_1 + 2 = 1 + 2 = 3$$ **Final answer:** $$R_1 = 1 \text{ ohm}, \quad R_2 = 3 \text{ ohms}$$