1. **State the problem:** We have a rectangle where the length is 5 cm greater than the width. The area of the rectangle is 84 cm². We need to find the length and the width. 2. **Formula used:** The area $A$ of a rectangle is given by: $$A = \text{length} \times \text{width}$$ 3. **Set variables:** Let the width be $w$ cm. Then the length is $w + 5$ cm. 4. **Write the equation for area:** $$w \times (w + 5) = 84$$ 5. **Expand and simplify:** $$w^2 + 5w = 84$$ 6. **Bring all terms to one side:** $$w^2 + 5w - 84 = 0$$ 7. **Solve the quadratic equation:** Use the quadratic formula: $$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=5$, and $c=-84$. Calculate the discriminant: $$\Delta = 5^2 - 4 \times 1 \times (-84) = 25 + 336 = 361$$ Calculate the roots: $$w = \frac{-5 \pm \sqrt{361}}{2} = \frac{-5 \pm 19}{2}$$ 8. **Find possible values for $w$:** - $w = \frac{-5 + 19}{2} = \frac{14}{2} = 7$ - $w = \frac{-5 - 19}{2} = \frac{-24}{2} = -12$ Since width cannot be negative, $w = 7$ cm. 9. **Find the length:** $$\text{length} = w + 5 = 7 + 5 = 12 \text{ cm}$$ **Final answer:** The width is 7 cm and the length is 12 cm.