Description for Correct answer:
I. \( \Large x^{2} \) - x - 6 = 0

=> \( \Large x^{2} \) - 3x + 2x - 6 = 0

=> x ( x - 3 ) + 2 ( x - 3 ) = 0

=> ( x + 2) (x - 3) = 0

=> x = -2 or 3

II. 2\( \Large y^{2} \) + 13y + 21 = 0

=> 2\( \Large y^{2} \) + 7y + 6y + 2l = 0

=> y ( 2y + 7 ) + 3 ( 2y + 7 ) = 0

=> ( y + 3 ) ( 2y + 7 ) = 0

\( \Large \therefore \) y = - 3 or \( \Large \frac{-7}{2} \)

Clearly, x > y