$ N(x) = x^2 + 5x -14 = 0$
$ \to x^2 + 7x - 2x - 14 = 0$
$\to x(x+7) -2(x+7) = 0$
$\to (x-2)(x+7) = 0$
$\to$ \(\left[ \begin{array}{l}x-2=0\\x+7=0\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}x=2\\x=-7\end{array} \right.\)
Vậy $ x \in \{ -7; 2 \}$