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51 > 0. Since the sum 14 is positive, the factors of 51 are both positive. The problem is that 51 = 3*17 and 3 + 17 = 20 ≠ 14.

But we are lucky that 14 = 7 + 7. So wee can use the difference of the squares such as (7 - A)(7 + A) = 49 - A2 = 51, where -A2 = 2 = -(i√2)2 .

Thus, the numbers are 7 - i√2 and 7 + i√2.

or algebraically,

x + y = 14; y = 14 - x

xy = 51

x(14 - x) = 51

-x2 +14x = 51

x2 -14x = -51

x2 -14x + 72 = -51 + 49

(x - 7)2 = -2

x - 7 = ± √-2

x = 7 ± i√2

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