To solve this system of equations, we can substitute the first equation into the second equation to eliminate one of the variables:

$$x^2 + y^2 = 13$$

$$(5 + 2y)^2 + y^2 = 13$$

$$25 + 10y + 4y^2 + y^2 = 13$$

$$25 + 14y + 5y^2 = 13$$

$$5y^2 + 14y - 12 = 0$$

This is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = 5$, $b = 14$, and $c = -12$. To solve this equation, we can use the quadratic formula:

$$y = \frac{-(14) \pm \sqrt{(14)^2 - 4(5)(-12)}}{2(5)}$$

This simplifies to:

$$y = \frac{-14 \pm \sqrt{196 + 240}}{10}$$

We can then simplify further to find the solutions:

$$y = \frac{-14 \pm \sqrt{436}}{10}$$

The solutions of this quadratic equation are:

$$y = \frac{-14 - 21.14}{10} \approx -3.57$$

$$y = \frac{-14 + 21.14}{10} \approx 2.64$$

Now that we have found the values of y, we can substitute them into the first equation to find the corresponding values of x:

$$x + y = 5 + 2y$$

$$x + (-3.57) = 5 + 2(-3.57)$$

$$x = -6.57 + 11.14 = 4.57$$

$$x + (2.64) = 5 + 2(2.64)$$

$$x = 5 + 5.28 = 10.28$$

Therefore, the solutions to the system of equations are (4.57, -3.57) and (10.28, 2.64).