Answer:
x ≤ 40 . . . miles per hour
Explanation:
You want to know the values of x for which d(x) = 0.05x² +x is at most 120.
Solution
d(x) ≤ 120 . . . . . . . . the required limit
0.05x² +x ≤ 120 . . . . . substitute the expression for d(x)
x² +20x -2400 ≤ 0 . . . . . multiply by 20, subtract 2400
(x -40)(x +60) ≤ 0 . . . . . . . . factor. Factors are zero for x=40, x=-60.
The product will be negative for x-values between the zeros:
-60 ≤ x ≤ 40
We only care about positive values of x:
0 ≤ x ≤ 40 . . . . . miles per hour