(x^2-4x+4)^2 +2(x-2)^2=3 Решите пожалуйста через замену переменных

$(x^2-4x+4)^2 +2(x-2)^2=3\\\\ \ [x^2-4x+4]^2 +2(x-2)^2=3\\\\ \ [(x-2)^2]^2 +2(x-2)^2-3=0\\\\ y=(x-2)^2 \geq 0\\\\ y^2+2y-3=0\\\\ y^2-y+3y-3=0\\\\ y(y-1)+3(y-1)=0\\\\ (y-1)(y+3)=0\\\\ y-1=0\ \ or\ \ y+3=0\\\\ y=1\ \ or\ \ y=-3\\\\ (x-2)^2=1\\\\ (x-2)-1^2=0\\\\ \ [(x-2)-1]*[(x-2)+1]=0\\\\ (x-3)(x-1)=0\\\\ x-3=0\ \ or\ \ x-1=0\\\\ x=3\ \ or\ \ x=1\\\\$

Ответ: $1;\ 3$