If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$ 
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$

$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{1-x}+1\right]$$
Now $$f(a) = \frac{a^2+1}{a^2+2} = \frac{a^2+2-1}{a^2+2} = 1-\frac{1}{a^2+2}$$
So $$f(x) = 1-\frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2}$$
Now put $1-x=t\;,$ Then we get $$f(t) =1- \frac{1}{\left(t-\frac{4}{t}+1\right)^2+2}$$
Now How can I maximize $\displaystyle \frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2, }\;,$ Help Required, Thanks

Write
$$
\frac{a^2+1}{a^2+2}=1-\frac{1}{a^2+2}
$$
Minimizing this is the same as maximizing $1/(a^2+2)$ which, in turn, is the same as minimizing $a^2+2$ or, as well, minimizing $a^2$.
Since
$$
a=-\frac{x^2-3x-2}{x-1}
$$
the minimum value for $a^2$ is obtained when $x^2-3x-2=0$.
