If the range of the function $f\left(x\right)=\frac{x^2+ax+b}{x^2+2x+3}$  is $[-5,4],a,b\in N$ , then find the value of $(a^2+b^2-2ab)$  _______

$\text{ Let }y=\frac{x^2+ax+b}{x^2+2x+3}$

$\Rightarrow x^2(1-y)+x(a-2y)+b-3y=0$

$D\ge 0$

$\Rightarrow \left(a^2+4y^2-4ay\right)-4\left(b-3y-by+3y^2\right)\ge 0$

$\begin{array}{r}\Rightarrow -8y^2-4y(a-3-b)+a^2-4b\ge 0\Rightarrow 8y^2+4y(a-3-b)+4b-a^2\le 0 ...\left(1\right)\end{array}$

Now compare with $(y+5)(y-4)\le 0$

$\begin{array}{r}{y}^2+y-20\le 0 ....\left(2\right)\end{array}$

$\frac{8}{1}=\frac{4(a-3-b)}{1}=\frac{4b-a^2}{-20}$

$\Rightarrow a^2+b^2-2ab=196+81-2\left(14\right)\left(9\right)=25$