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The two roots given are x = 3+2i and x = 3-2i. Therefore x - (3+2i) = 0, and x - (3-2i) = 0. This also implies that [x - (3+2i)]*[x - (3-2i)] = 0. Rewriting: (x - 3 - 2i)(x - 3 + 2i) = 0 Multiplying out: x^2 - 3x + 2ix - 3x + 9 - 6i - 2ix + 6i - 4i^2 = 0 We note that the 2ix's cancel, as well as the 6i's: x^2 - 3x - 3x + 9 - 4i^2 = 0 And finally, noting that i^2 = -1, we combine like terms and get: x^2 - 6x + 13 = 0, which is the required quadratic equation.

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