\(\Delta=\left(m+2\right)^2-4m^2-4=4m-3m^2\ge0\Rightarrow0\le m\le\frac{4}{3}\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=m+2\\x_1x_2=m^2+1\end{matrix}\right.\)
\(P=4\left(x_1+x_2\right)-x_1x_2=4\left(m+2\right)-\left(m^2+1\right)\)
\(P=-m^2+4m+7\)
Xét trên đoạn \(\left[0;\frac{4}{3}\right]\) ta có: \(P\left(0\right)=7\); \(P\left(\frac{4}{3}\right)=\frac{95}{9}\)
\(\Rightarrow P_{max}=\frac{95}{9}\) khi \(m=\frac{4}{3}\)