If domain of the function ${\log }_e\left(\frac{6x^2+5x+1}{2x-1}\right)+{\cos }^{-1}\left(\frac{2x^2-3x+4}{3x-5}\right)$ is $(\alpha ,\beta )\cup (\gamma ,\delta ),$ then $\frac{18({\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2)}{2}$ is equal to ___________

We need to find the domain of the function  ${\log }_e\left(\frac{6x^2+5x+1}{2x-1}\right)+{\cos }^{-1}\left(\frac{2x^2-3x+4}{3x-5}\right)$
 $\Rightarrow \frac{6x^2+5x+1}{2x-1}>\mathrm{0.....}\left(1\right)\text{ and }\Rightarrow -1\le \frac{2x^2-3x+4}{3x-5}\le \mathrm{1.....}\left(2\right)$
Now from $\left(1\right)$ , 
 $\Rightarrow \frac{6x^2+5x+1}{2x-1}>0\Rightarrow \frac{(3x+1)(2x+1)}{(2x-1)}>0$
Now we get the common region here as 
 $\Rightarrow x\in (-\frac{1}{2},-\frac{1}{3})\cup (\frac{1}{2},\infty )\mathrm{....}\left(a\right)$
Now from $\left(2\right)$ we get that
 $\Rightarrow -1\le \frac{2x^2-3x+4}{3x-5}\le 1$
 $\Rightarrow \frac{2x^2-3x+4}{3x-5}\ge -\mathrm{1.....}\left(3\right)$
And  $\frac{2x^2-3x+4}{3x-5}\le \mathrm{1.....}\left(4\right)$
From $\left(3\right)$ we get  $\Rightarrow \frac{2x^2-3x+4}{3x-5}+1\ge 0$
 $\Rightarrow x\in [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]\cup (\frac{5}{3},\infty )\mathrm{......}\left(b\right)$
From $\left(4\right)$ we get 
 $\Rightarrow \frac{2x^2-3x+4}{3x-5}-1\le 0\Rightarrow \frac{2x^2-6x+9}{3x-5}\ge 0$
 $\text{As }D={(-6)}^2-4\left(2\right)\left(9\right)<0\text{ and }2>0$
We observe that  $2x^2-6x+9>0\forall x\in R$
 $\Rightarrow \frac{1}{3x-5}\le 0(\because 2x^2-6x+9)>0\forall x\in R$
 $\Rightarrow x\in (-\infty ,\frac{5}{3})\mathrm{....}\left(c\right)$
$\Rightarrow$ Intersection of $\left(a\right),\left(b\right)\text{ and }\left(c\right)$ gives us   $\Rightarrow x\in (-\frac{1}{2},-\frac{1}{3})\cup (\frac{1}{2},\frac{1}{\sqrt{2}})$
On comparing this with  $\Rightarrow (\alpha ,\beta )\cup (\gamma ,\delta ),$
 $\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2=\frac{10}{9}$
 $\Rightarrow 18({\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2)=20$
Hence this is the required answer.