If x  is real then the maximum value of  $\frac{x^2+14x+9}{x^2+2x+3}$ is 

$$\begin{gathered} Let y=\frac{x^2+14x+9}{x^2+2x+y}\Rightarrow (y-1)x^2+(2y-14)x+(3y-9)=0 \\ If x is real then {(2y-14)}^2-4(y-1)(3y-9)\ge 0 \end{gathered}$$

$\begin{array}{l}\Rightarrow y^2+y-20\le 0\Rightarrow (y+5)(y-4)\le 0\\ \Rightarrow -5\le y\le 4\end{array}$

Maximum value =4