Roots of a quadratic equation

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If 2 and 3/2 are the roots of a second degree polynomial P(y) we could have written the polynomial as:

P(y)=(y-2)(y-3/2)

which gives:

P(y) = y^2-2y-3y/2+3

P = (y^2) - (7y/2) + 3

And this is NOT the original polynomial:

Q(y) = 2y^2 - 7y + 3

Thus, 2 and 3/2 are not the roots of Q(x).

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