\( \Large x^{2} - x - 12 = 0 \)

=\( \Large x^{2} - 4x + 3x - 12 = 0 \)

=\( \Large x \left(x-4\right) + 3 \left(x-4\right) = 0 \)

=\( \Large \left(x-4\right) \left(x+3\right) = 0 \)

Therefore, x = -3, 4

and \( \Large y^{2} + 5y + 6 = 0 \)

=\( \Large y^{2} + 3y + 2y + 6 = 0 \)

=\( \Large y \left(y+3\right) + 2 \left(y+3\right) = 0 \)

=\( \Large \left(y + 3\right) \left(y - 3\right) = 0 \)

= y = -3, -2

Therefore, \( \Large x \ge y \)

[because x = -3 and y = -3, so x = y and x = 4 and y = -2, hence x > y]