If ${(x^2+x+2)}^2-(a-3)(x^2+x+1)(x^2+x+2)+(a-4){(x^2+x+1)}^2=0$  has at least one root then the complete set of values of a.

Let  $t=x^2+x+1\Rightarrow t\in [\frac{3}{4},\infty )$
Hence,  ${(t+1)}^2-(a-3)t(t+1)+(a-4)t^2=0$
$\Rightarrow t^2+2t+1-(a-3)(t^2+t)+(a-4)t^2=0$ 
 $\Rightarrow t(2-a+3)+1=0\Rightarrow t=\frac{1}{a-5}\Rightarrow \frac{1}{a-5}\ge \frac{3}{4} \therefore a\in (5,\frac{19}{3}]$