X^2-4X+19=0

This doesn't have any real solutions....

Here are the non-real ones

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{19}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{15}}}}{\mathtt{\,\times\,}}{i}\\
{\mathtt{x}} = {\sqrt{{\mathtt{15}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{3.872\: \!983\: \!346\: \!207\: \!416\: \!9}}{i}\\
{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.872\: \!983\: \!346\: \!207\: \!416\: \!9}}{i}\\
\end{array} \right\}$$

19 is a prime number so you will have to use the quadratic formula to solve this one.