<s>[INST] Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
 [/INST] First you need to factor before bring the terms into one side. $$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=(2x-3)^2$$ Now you subtract the right hand side, where it becomes $$(x^2-3x+1)^2-(2x-3)^2=0$$ You can factor by using differences of squares where $a^2-b^2=(a+b)(a-b)$. 
Now finish this off on your own:)
If you really want the full solution just mouse over this part

 $$(x^2-3x+1)^2-(2x-3)^2=0 $$ $$(x^2-x-2)(x^2-5x+4)=0$$ $$(x-2)(x+1)(x-4)(x-1)=0$$ $$x=-1,1,2,4$$

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