I. 3x2 - 16x - 12 = 0
II. 2y2 + 11y + 9 = 0
I. 3x2 - 16x - 12 = 0 => 3x2 - 18x + 2x - 12 = 0 => 3x(x - 6) + 2(x - 6) = 0 => (x - 6) (3x + 2) = 0 => x = 6, -2/3 II. 2y2 + 11y + 9 = 0 => 2y2 + 9y + 2y + 9 = 0 => y (2y + 9) + 1(2y + 9) = 0 => (y + 1) (2y + 9) = 0 => y = -1, -9/2 Hence, x > y
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