a) 2x ( $x^{2}$ - 4 ) = 0
⇔2x (x-2) (x+2) = 0
⇔\(\left[ \begin{array}{l}x=0\\x-2=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=2\\x=-2\end{array} \right.\)
b) 2 ( x+5 ) - $x^{2}$ - 5x = 0
⇔2 (x+5) - x (x+5) = 0
⇔(2-x) (x+5) = 0
⇔\(\left[ \begin{array}{l} 2-x=0\\x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l} x=2\\x=-5\end{array} \right.\)