\(\Delta'=\left(m+3\right)^2-\left(m^2+3\right)=6m+6>0\Rightarrow m>-1\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+3\right)\\x_1x_2=m^2+3\end{matrix}\right.\)
\(\Rightarrow2\left(m+3\right)=m^2+3\)
\(\Leftrightarrow m^2-2m-3=0\Rightarrow\left[{}\begin{matrix}m=-1\left(l\right)\\m=3\end{matrix}\right.\)