Find the least positive integer such that its square is greater than 5 times of the integer by –6.

Correct option is D

Given:
​$$x^2 > 5x - 6$$​​
Solution:
​$$x^2 > 5x - 6 \\x^2 - 5x + 6 > 0\\x^2 - 5x + 6 = (x - 2)(x - 3)$$​​
So, the inequality becomes:
(x - 2)(x - 3) > 0
​least positive integer is 2.

Thus, Option (D) is right.