<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] (x$^2$ - 3x + 1)$^2$ = (2x - 3)$^2$
(x$^2$ - 3x + 1)$^2$ - (2x - 3)$^2$ = 0
(x$^2$-3x+1-2x+3)(x$^2$-3x+1+2x-3) = 0
(x$^2$-5x+4)(x$^2$-x-2) = 0
(x-1)(x-4)(x+1)(x-2) = 0
Therefore, solution is x = 1, x = 4, x = -1 or x = 2
</s>