If 2 is a root then substituting the value 2 in place of x should satisfy the equation.

A. Given, x² - 4x + 5 = 0

⇒ (2)² - 4(2) + 5 = 1 ≠ 0

So, x = 2 is not a root of x² - 4x + 5 = 0

B. Given, x² + 3x – 12 = 0

⇒ (2)² + 3(2) – 12 = -2 ≠ 0

So, x = 2 is not a root of x² + 3x – 12 = 0

C. Given, 2x² - 7x + 6 = 0

⇒ 2(2)² - 7(2) + 6 = 0

Here, x = 2 is a root of 2x² - 7x + 6 = 0

D. Given, 3x² - 6x – 2 = 0

⇒ 3(2)² - 6(2) – 2 = -2 ≠ 0

So, x = 2 is not a root of 3x² - 6x – 2 = 0