$16-{x}^{2}=0$
Use ${a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right)$ to factor the expression$\left( 4-x \right) \times \left( 4+x \right)=0$
When the product of factors equals $0$, at least one factor is $0$$\begin{array} { l }4-x=0,\\4+x=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=4,\\4+x=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=4,\\x=-4\end{array}$
The equation has $2$ solutions$\begin{array} { l }x_1=-4,& x_2=4\end{array}$