Solution of $$\dfrac{x - 2}{x + 5} > 2$$​ is:

Given inequality:
​$$\dfrac{x - 2}{x + 5} > 2$$​​
Solution:
​$$\dfrac{x - 2}{x + 5} - 2 > 0 \\ \ \\\Rightarrow \dfrac{x - 2 - 2(x + 5)}{x + 5} > 0 \\ \ \\\Rightarrow \dfrac{x - 2 - 2x - 10}{x + 5} > 0 \\ \ \\\Rightarrow \dfrac{-x - 12}{x + 5} > 0$$​​
Multiply numerator and denominator by -1 to simplify:

​$$\Rightarrow \dfrac{x + 12}{x + 5} < 0$$​​
Find critical points
Set numerator and denominator equal to 0.
x + 12 = 0 $$\Rightarrow x = -12$$​​
x + 5 = 0 $$\Rightarrow x = -5$$​​
These divide the number line into intervals:
​$$(-\infty, -12), (-12, -5), (-5, \infty)$$​​
Sign analysis of $$f(x) = \dfrac{x + 12}{x + 5}:$$​​
For x > -5:
Choose  $$x = 0 \Rightarrow f(0) = \dfrac{12}{5} > 0$$​ (Positive)
For $$-12 < x < -5:$$​​
Choose $$x = -10 \Rightarrow f(-10) = \dfrac{-10 + 12}{-10 + 5} = \dfrac{2}{-5} < 0$$​ (Negative)
For x < -12:
Choose $$x = -13 \Rightarrow f(-13) = \dfrac{-13 + 12}{-13 + 5} = \dfrac{-1}{-8} > 0$$​ (Positive)

We need $$\dfrac{x + 12}{x + 5} < 0$$​, which occurs for
​$$-12 < x < -5$$​​
Therefore, the solution is:
​$$x \in (-12, -5)$$​​
\textbf{Correct answer: (c) $$x \in (-12, -5)$$​​