Identifying the two consecutive odd positive integers
To find the two consecutive odd positive integers whose sum of squares is 290, let's assume the first odd integer is \(x\) and the next consecutive odd integer is \(x + 2\).

Formulating the equation
According to the given condition, the sum of the squares of these two integers is 290:
\[ x^2 + (x + 2)^2 = 290 \]

Solving the equation
Expanding the equation and simplifying, we get:
\[ x^2 + x^2 + 4x + 4 = 290 \]
\[ 2x^2 + 4x + 4 = 290 \]
\[ 2x^2 + 4x - 286 = 0 \]
Now, we can solve this quadratic equation to find the value of \(x\).

Calculating the values
Using the quadratic formula, we find:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values of \(a = 2\), \(b = 4\), and \(c = -286\), we get:
\[ x = \frac{-4 \pm \sqrt{4^2 - 4*2*(-286)}}{2*2} \]
\[ x = \frac{-4 \pm \sqrt{16 + 2288}}{4} \]
\[ x = \frac{-4 \pm \sqrt{2304}}{4} \]
\[ x = \frac{-4 \pm 48}{4} \]
Therefore, the two consecutive odd positive integers are \(7\) and \(9\), as their sum of squares is indeed 290.