`f(x) = {{:((2x^(2)-3x-2)/(x-2), if x ne 2), (5, if x = 2):}` at `x = 2`.

Explanation

To determine whether the function \( f(x) \) is continuous at \( x = 2 \), we need to check the left-hand limit, right-hand limit, and the value of the function at that point.

Given:
\[ f(x) = \begin{cases} \frac{2x^2 - 3x - 2}{x - 2} & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit as \( x \) approaches 2
We calculate the left-hand limit:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2x^2 - 3x - 2}{x - 2} \]

Substituting \( x = 2 - h \) where \( h \to 0^+ \):
\[ \lim_{h \to 0} \frac{2(2 - h)^2 - 3(2 - h) - 2}{(2 - h) - 2} \]

Expanding the numerator:
\[ = \lim_{h \to 0} \frac{2(4 - 4h + h^2) - 6 + 3h - 2}{-h} \]
\[ = \lim_{h \to 0} \frac{8 - 8h + 2h^2 - 6 + 3h - 2}{-h} \]
\[ = \lim_{h \to 0} \frac{2h^2 - 5h}{-h} \]
\[ = \lim_{h \to 0} - (2h - 5) \]
As \( h \to 0 \):
\[ = 5 \]

### Step 2: Calculate the Right-Hand Limit as \( x \) approaches 2
Now we calculate the right-hand limit:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{2x^2 - 3x - 2}{x - 2} \]

Substituting \( x = 2 + h \) where \( h \to 0^+ \):
\[ \lim_{h \to 0} \frac{2(2 + h)^2 - 3(2 + h) - 2}{(2 + h) - 2} \]

Expanding the numerator:
\[ = \lim_{h \to 0} \frac{2(4 + 4h + h^2) - 6 - 3h - 2}{h} \]
\[ = \lim_{h \to 0} \frac{8 + 8h + 2h^2 - 6 - 3h - 2}{h} \]
\[ = \lim_{h \to 0} \frac{2h^2 + 5h}{h} \]
\[ = \lim_{h \to 0} (2h + 5) \]
As \( h \to 0 \):
\[ = 5 \]

### Step 3: Evaluate \( f(2) \)
From the definition of the function:
\[ f(2) = 5 \]

### Conclusion
Since:
\[ \lim_{x \to 2^-} f(x) = 5, \quad \lim_{x \to 2^+} f(x) = 5, \quad \text{and } f(2) = 5 \]
We conclude that:
\[ \lim_{x \to 2} f(x) = f(2) \]
Thus, \( f(x) \) is continuous at \( x = 2 \).