\(\Delta'=\left(m+4\right)^2-\left(m^2-8\right)=m^2+8m+16-m^2+8=8m+24\ge0\)
\(\Leftrightarrow m\ge-3\)
Theo Viete: \(\left\{{}\begin{matrix}x_1+x_2=2m+8\\x_1x_2=m^2-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=\frac{x_1+x_2-8}{2}\\x_1x_2=\left(\frac{x_1+x_2-8}{2}\right)^2-8\end{matrix}\right.\)
\(\Leftrightarrow x_1^2+x_2^2-16x_1-16x_2-2x_1x_2+32=0\)