1. **State the problem:** Wilbur has a rectangular pool 12 feet by 25 feet and wants to build a concrete walkway of equal width $x$ around it. The total area of the pool plus walkway is 475 square feet. We need to find the width $x$ of the walkway. 2. **Set up the equation:** The dimensions of the pool plus walkway are $(25 + 2x)$ by $(12 + 2x)$ because the walkway adds $x$ feet on each side. 3. **Write the area equation:** $$ (25 + 2x)(12 + 2x) = 475 $$ 4. **Expand the left side:** $$ 25 \times 12 + 25 \times 2x + 2x \times 12 + 2x \times 2x = 475 $$ $$ 300 + 50x + 24x + 4x^2 = 475 $$ 5. **Combine like terms:** $$ 4x^2 + 74x + 300 = 475 $$ 6. **Bring all terms to one side:** $$ 4x^2 + 74x + 300 - 475 = 0 $$ $$ 4x^2 + 74x - 175 = 0 $$ 7. **Solve the quadratic equation using the quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=4$, $b=74$, and $c=-175$. 8. **Calculate the discriminant:** $$ \Delta = 74^2 - 4 \times 4 \times (-175) = 5476 + 2800 = 8276 $$ 9. **Calculate the square root:** $$ \sqrt{8276} \approx 90.98 $$ 10. **Find the two possible solutions:** $$ x = \frac{-74 \pm 90.98}{8} $$ 11. **Calculate each:** - Positive root: $$ x = \frac{-74 + 90.98}{8} = \frac{16.98}{8} = 2.1225 $$ - Negative root: $$ x = \frac{-74 - 90.98}{8} = \frac{-164.98}{8} = -20.6225 $$ 12. **Interpret the result:** Width cannot be negative, so the walkway width is approximately $2.12$ feet. **Final answer:** The width of the walkway should be approximately $\boxed{2.12}$ feet.