The two zeroes of a quadratic polynomial add up to −6, and the sum of their squares is 18.
Is such a polynomial possible, and if so what are its zeroes?

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$ gives $18=36-2\alpha \beta$, so $\alpha \beta =9$. Then $(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta =36-36=0$, so the zeroes are equal, and each is $\frac{-6}{2}=-3$. The polynomial is $x^2+6x+9$. The pair −3 and 3 adds to 0, not −6; the pair 0 and −6 has sum of squares 36, not 18; and the product works out to +9, so nothing forces it to be negative.