Đặt $x^2+x-2=a$
Khi đó $A=(a-4).(a+4)$
$=a^2-16$
Mà $a^2 \geq 0⇒a^2-16≥-16$
Hay $A≥-16$
Dấu $=$ xảy ra $⇔a^2=0⇔a=0⇔x^2+x-2=0$
$⇔x^2-x+2x-2=0$
$⇔(x-1)(x+2)=0$
$⇔$\(\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)