Is there another way to solve this quadratic equation? $$\frac { 4 }{ x^{ 2 }-2x+1 } +\frac { 7 }{ x^{ 2 }-2x+4 } =2$$
Steps I took:
$$\frac { 4(x^{ 2 }-2x+4) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } +\frac { 7(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } =\frac { 2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } $$
$$4(x^{ 2 }-2x+4)+7(x^{ 2 }-2x+1)=2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1)$$
$$11x^{ 2 }-22x+23=2x^{ 4 }-8x^{ 3 }+18x^{ 2 }-20x+8$$
I can keep going with all the steps I took, but is there a more elegant way to arrive at the solution for this equation? It seems as if I keep going the way I am, I will hit a dead end. No actual solution, please. Hints are much better appreciated.  

By inspection, the two denominators differ by $3$, just as the numerators do, hinting that to establish
$$\frac44+\frac77=2,$$ we may set
$$x^2-2x+1=(x-1)^2=4,$$ i.e. $$\color{green}{x=-1\lor x=3}.$$
The other two roots are a little more elusive. 
But we can observe that when expanding
$$\frac4z+\frac7{z+3}-2=0,$$ we will get terms $-2z^2$ and $12$, so that the product of the $z$ roots is $-6$, and
$$(x-1)^2=-\frac32,$$ i.e. $$\color{green}{x=1-i\sqrt{\frac32}\lor x=1+i\sqrt{\frac32}}.$$
