We can use the distance formula to find the lengths of the sides of the triangle.
The distance between (2,2) and (5,6) is $\sqrt{(5-2)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The distance between (5,6) and (6,2) is $\sqrt{(6-5)^2 + (2-6)^2} = \sqrt{1 + 16} = \sqrt{17}$.
The distance between (6,2) and (2,2) is $\sqrt{(2-6)^2 + (2-2)^2} = \sqrt{16 + 0} = \sqrt{16} = 4$.
Therefore, the longest side of the triangle has length $\boxed{5}$ units.
The answer is: 5
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
The triangle with vertices at (2,2), (5,6), and (6,2) has sides of varying lengths. What is the length, in units, of the longest side of the triangle?

### Answer:
