To solve the system of equations using the elimination method, we start with the two given equations:
- \( y = x^2 - 2x - 2 \) (Equation 1)
- \( y = 2x - 6 \) (Equation 2)
Since both equations are equal to \( y \), we can set them equal to each other:
\[ x^2 - 2x - 2 = 2x - 6 \]
Now, let's rearrange this equation to one side:
\[ x^2 - 2x - 2 - 2x + 6 = 0 \] \[ x^2 - 4x + 4 = 0 \]
Next, we can factor this quadratic equation:
\[ (x - 2)^2 = 0 \]
This gives us the solution:
\[ x - 2 = 0 \implies x = 2 \]
Now that we have \( x = 2 \), we can substitute this value back into either of the original equations to find \( y \). We will use Equation 2:
\[ y = 2(2) - 6 = 4 - 6 = -2 \]
Thus, we have one solution:
\[ (2, -2) \]
To verify if there are any other solutions, we can substitute back into the first equation as well.
When we have a quadratic and we found a value of \( x \) that gives a unique \( y \) value, we need to check if there might be any other intersections. The quadratic may have multiple real solutions, but they manifest as the same intersection point in this case.
Since the quadratic factors perfectly, it confirms that \( x = 2 \) is the only solution derived from this equation.
Therefore, the only solution to the system is:
\[ \boxed{(2, -2)} \]
This corresponds to option B.