Marcus can walk 1 mile to the library in the same time he can bicycle 2.5 miles to the sporting goods store. His speed on the bicycle is 6 mi/h faster than his walking speed. What is his speed walking?
Let Marcus' speed walking be x mi/h. His speed biking is x + 6 mi/h.
Since time = distance/speed, the time it takes him to walk 1 mile is 1/x, and the time it takes him to bike 2.5 miles is 2.5/(x + 6).
Setting these two expressions equal to each other, we have 1/x = 2.5/(x + 6).
Cross multiplying, we get x(x + 6) = 2.5.
Expanding the left side of the equation, we have x^2 + 6x = 2.5.
Rearranging the equation, we have x^2 + 6x - 2.5 = 0.
Using the quadratic formula, we have x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = 6, and c = -2.5.
Calculating, we have x = (-6 ± √(6^2 - 4(1)(-2.5))) / (2(1)).
Simplifying the expression, we have x = (-6 ± √(36 + 10)) / 2.
Further simplifying, we have x = (-6 ± √46) / 2.
Thus, x = (-6 + √46) / 2 and x = (-6 - √46) / 2 are the possible values for x.
Calculating these values, we have x = (-6 + √46) / 2 ≈ 1.68 and x = (-6 - √46) / 2 ≈ -3.68.
Since we're looking for a positive speed, we can conclude Marcus' speed walking is approximately 1.68 mi/h. Answer: \boxed{1.68}.