1. **State the problem:** We are given a rectangle with area 28 yd². The length is 1 yd more than twice the width. We need to find the length and width. 2. **Set variables:** Let the width be $w$ yd. 3. **Express length in terms of width:** Length $L = 2w + 1$ yd. 4. **Write the area formula:** Area $A = L \times w$. 5. **Substitute known values:** $28 = (2w + 1) \times w$. 6. **Form quadratic equation:** $28 = 2w^2 + w$. 7. **Rewrite equation:** $2w^2 + w - 28 = 0$. 8. **Use quadratic formula:** For $ax^2 + bx + c = 0$, $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=2$, $b=1$, $c=-28$. 9. **Calculate discriminant:** $\Delta = 1^2 - 4 \times 2 \times (-28) = 1 + 224 = 225$. 10. **Calculate roots:** $$w = \frac{-1 \pm \sqrt{225}}{2 \times 2} = \frac{-1 \pm 15}{4}$$ 11. **Evaluate roots:** - $w = \frac{-1 + 15}{4} = \frac{14}{4} = 3.5$ - $w = \frac{-1 - 15}{4} = \frac{-16}{4} = -4$ (discard negative width) 12. **Find length:** $$L = 2 \times 3.5 + 1 = 7 + 1 = 8$$ 13. **Final answer:** - Width = 3.5 yd - Length = 8 yd