I. 2x2 + 12x + 18 = 0
II. 3y2 + 13y + 12 = 0
I. 2x2 + 12x + 18 = 0 => 2x2 + 6x + 6x + 18 = 0 => 2x(x + 3) + 6(x + 3) = 0 => (x + 3) (2x + 6) = 0 => x = -3, -3 II. 3y2 + 13y + 12 = 0 => 3y2 + 9y + 4y + 12 = 0 => 3y(y + 3) + 4(y + 3) = 0 => (y + 3) (y + 4) = 0 => y = -3, -4 Hence, x ≥ y.
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