Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$. What is $x+y$?

$ \textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8 $
1. Start with the given equation:
   \[
   x^2 + y^2 = 10x - 6y - 34
   \]

2. Rearrange the equation to group the $x$ and $y$ terms on one side:
   \[
   x^2 - 10x + y^2 + 6y = -34
   \]

3. Complete the square for the $x$ terms. To complete the square for $x^2 - 10x$, add and subtract $(\frac{10}{2})^2 = 25$:
   \[
   x^2 - 10x + 25 - 25
   \]
   This can be written as:
   \[
   (x - 5)^2 - 25
   \]

4. Complete the square for the $y$ terms. To complete the square for $y^2 + 6y$, add and subtract $(\frac{6}{2})^2 = 9$:
   \[
   y^2 + 6y + 9 - 9
   \]
   This can be written as:
   \[
   (y + 3)^2 - 9
   \]

5. Substitute these completed squares back into the equation:
   \[
   (x - 5)^2 - 25 + (y + 3)^2 - 9 = -34
   \]

6. Combine the constants on the right-hand side:
   \[
   (x - 5)^2 + (y + 3)^2 - 34 = -34
   \]

7. Add 34 to both sides to isolate the squares:
   \[
   (x - 5)^2 + (y + 3)^2 = 0
   \]

8. Since the sum of two squares is zero, each square must individually be zero:
   \[
   (x - 5)^2 = 0 \quad \text{and} \quad (y + 3)^2 = 0
   \]

9. Solve for $x$ and $y$:
   \[
   x - 5 = 0 \quad \Rightarrow \quad x = 5
   \]
   \[
   y + 3 = 0 \quad \Rightarrow \quad y = -3
   \]

10. Find $x + y$:
    \[
    x + y = 5 + (-3) = 2
    \]

The final answer is $\boxed{2}$