Let $a \spadesuit b = \frac{a^2-b^2}{2b-2a}$ . Given that $3 \spadesuit x = -10$, compute $x$.
1. Given the operation \( a \spadesuit b = \frac{a^2 - b^2}{2b - 2a} \), we need to solve for \( x \) in the equation \( 3 \spadesuit x = -10 \).

2. Substitute \( a = 3 \) and \( b = x \) into the operation:
   \[
   3 \spadesuit x = \frac{3^2 - x^2}{2x - 2 \cdot 3} = \frac{9 - x^2}{2x - 6}
   \]

3. Set the expression equal to \(-10\):
   \[
   \frac{9 - x^2}{2x - 6} = -10
   \]

4. To eliminate the fraction, multiply both sides by \( 2x - 6 \):
   \[
   9 - x^2 = -10(2x - 6)
   \]

5. Distribute \(-10\) on the right-hand side:
   \[
   9 - x^2 = -20x + 60
   \]

6. Rearrange the equation to form a standard quadratic equation:
   \[
   x^2 - 20x + 51 = 0
   \]

7. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -20 \), and \( c = 51 \):
   \[
   x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 51}}{2 \cdot 1} = \frac{20 \pm \sqrt{400 - 204}}{2} = \frac{20 \pm \sqrt{196}}{2} = \frac{20 \pm 14}{2}
   \]

8. This gives us two potential solutions:
   \[
   x = \frac{20 + 14}{2} = 17 \quad \text{and} \quad x = \frac{20 - 14}{2} = 3
   \]

9. Check the solutions to ensure they are valid. If \( x = 3 \), the denominator \( 2x - 6 \) becomes zero, which is undefined. Therefore, \( x = 3 \) is not a valid solution.

10. The only valid solution is \( x = 17 \).

The final answer is \( \boxed{17} \)