I. x2 - 5x - 14 = 0
II. y2 - 16y + 64 = 0
I. x2 - 5x - 14 = 0 => x2 - 7x + 2x - 14 = 0 => x(x - 7) + 2(x - 7) = 0 => (x - 7) (x + 2) = 0 => x = 7, -2 II. y2 - 16y + 64 = 0 => y2 - 8y - 8y + 64 = 0 => y(y - 8) - 8(y - 8) = 0 => (y - 8) (y - 8) = 0 => y = 8, 8 Hence, x < y.
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