Not so lengthy

Let $y^2-4y+5=\color{#20A900}{k}$ . Now observe that The Discriminant of the quadratic ( $\color{#20A900}{k}$ ) is $<0$ and the coefficient of $y^2$ which is $1 >0$ , which means the quadratic equation is always positive for $\forall y \in R$ .Now , we calculate the Infinite Sum ,

$\large \displaystyle a\sum_{r=0}^{\infty}\dfrac1{\color{#20A900}k^{r-1}} \implies a\times \dfrac{\color{#20A900}k}{1-\dfrac1{\color{#20A900}k}} \implies \dfrac{a\color{#20A900}k^2}{\color{#20A900}k-1}$

For the Infinite Sum to be meaningful, -1<\dfrac{1}{\color{#20A900}k}<1 \implies -\color{#20A900}k<1<\color{#20A900}k \implies 0<\color{#20A900}k-1 \text { & } 0<\color{#20A900}k+1

$\implies y^2-4y+5-1>0 \\ y^2-4y+4>0 \implies (y-2)^2>0 \\ \implies y\neq 2 \\ \text{Which Means } x^2-6x+11\neq 2 \\ x^2-6x+9\neq 0 \\ \implies \color{#3D99F6}{\boxed{x\neq 3}}$