$Q(x) = - x^2 - 5x + 6 = 0$
$\to x^2 + 5x -6 = 0$
$\to x^2 - x +6x - 6 = 0$
$\to x(x-1) + 6(x-1) = 0$
$\to (x+6)(x-1) = 0$
$\to $ \(\left[ \begin{array}{l}x+6=0\\x-1=0\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}x=-6\\x=1\end{array} \right.\)
Vậy $ x \in \{ 1; -6 \}$