x2+4x-16=0

x^2 - 4x + 16 = 0    this won't factor.....using the on-site solver, we have

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{16}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}\\
{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{3.464\: \!101\: \!615\: \!137\: \!754\: \!6}}{i}\\
{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.464\: \!101\: \!615\: \!137\: \!754\: \!6}}{i}\\
\end{array} \right\}$$

Note that neither solution is "real"......we could have seen this by using the "discriminant" - (b^2 - 4ac) - where b =-4, a = 1 and c = 16

So we have......  (-4)^2 - 4(1)(16)  = 16 - 64  = -48

And since this is < 0, we will have  "non-real"  solutions...