А как уравнение (x^2 - 2x + 4)^2 + (x-1)^2 = 27

Решите задачу:

$(x^2 - 2x + 4)^2 + (x-1)^2 = 27\\(x^2-2x+4)^2+x^2-2x+1=27\\(x^2-2x+4)^2+x^2-2x+4=30;x^2-2x+4=a\\a^2+a-30=0\\\begin{cases}a_1+a_2=-1\\a_1*a_2=-30\end{cases}\\a_1=-6\ a_2=5\\x^2-2x+4=-6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2-2x+4=5\\x^2-2x+10=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2-2x-1=0\\x_{1,2}=1^+_-\sqrt{-9}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{1,2}=1^+_-\sqrt{2}\\x\in\varnothing\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_1=1+\sqrt2\ \ x_2=1-\sqrt2$