Introduction:
In this problem, we are required to find two consecutive odd positive integers whose sum of the squares is equal to 290. Let's solve this problem step by step.

Step 1: Understanding the problem
To solve the problem, we need to understand the important information given in the problem statement. We are given:
- We need to find two consecutive odd positive integers.
- The sum of the squares of these integers is equal to 290.

Step 2: Formulating equations
Let's assume that the first odd integer is x. As the integers are consecutive odd integers, the next odd integer will be x + 2. Now, we can write the equation based on the given information as:

x^2 + (x + 2)^2 = 290

Expanding the equation, we get:

x^2 + x^2 + 4x + 4 = 290

Simplifying the equation, we get:

2x^2 + 4x - 286 = 0

Dividing both sides by 2, we get:

x^2 + 2x - 143 = 0

Step 3: Solving the equation
We can solve the quadratic equation using the quadratic formula:

x = [-b ± √(b^2 - 4ac)] / 2a

Substituting the values in the formula, we get:

x = [-2 ± √(2^2 - 4(1)(-143))] / 2(1)

x = [-2 ± √(4 + 572)] / 2

x = [-2 ± √576] / 2

x = [-2 ± 24] / 2

x = -13 or 11

As we need to find positive integers, we will take x = 11.

Step 4: Finding the consecutive odd integer
The first odd integer is x = 11. The next odd integer will be x + 2 = 13.

Step 5: Checking the answer
We can check whether our answer is correct or not by substituting the values in the equation:

11^2 + 13^2 = 121 + 169 = 290

The sum of the squares of these integers is equal to 290, which is the same as given in the problem statement.

Conclusion:
Therefore, we have found that the two consecutive odd positive integers whose sum of squares is equal to 290 are 11 and 13.