Just a Poly problem

$\begin{aligned} 2x^2 + \frac 1{3x^2-6x+9} & = \frac 1{x^2-6x+10} + 1 \\ 2x^2 - 1 + \frac 1{2x^2 -1+x^2-6x+10} & = \frac 1{x^2-6x+10} & \small \color{#3D99F6} \text{Let }a = 2x^2-1 \text{ and }b = x^2-6x+10 \\ \implies a + \frac 1{a+b} & = \frac 1b \\ \implies a & = 0 \\ 2x^2 - 1 & = 0 \\ \implies x & = \frac 1{\sqrt 2} \approx \boxed {0.7071} & \small \color{#3D99F6} \text{Since }x > 0 \end{aligned}$

$Let~a=2x^2,~~and~~b=x^2-6x+10.\\ Then~given~equation~is:-\\ a+1/(a+b-1)=1/b+1.\\ See~easily~a=1~is~the~solution.\\ \implies~2x^2=1,~~for~x>0, ~~x=\sqrt2/2=0.7071.$

If you try to compare L.H.S and R.H.S, we simply see that $x=\dfrac{1}{\sqrt{2}}$

First set $2x^2=1$ , which which will give you $x=\dfrac{1}{\sqrt{2}}$ (not a negative result because it's not allowed) and then set $\dfrac { 1 }{ 3x^{ 2 }-6x+9 }=\dfrac { 1 }{ x^{ 2 }-6x+10 }$ , which again gives you $x=\boxed{\dfrac{1}{\sqrt{2}}\approx 0.7071}$ , hence it satisfies!

Although It's not a solution...