$7)$
$$(x^2-4x+1)^2-3(x^2-4x+1)+1 = 2x$$
$$[(x^2-4x+1)^2-2(x^2-4x+1)+1]-[(x^2-4x+1)+2x] = 0$$
$$[(x^2-4x+1)-1]^2-(x^2-2x+1) = 0$$
$$(x^2-4x)^2-(x-1)^2 = 0$$
$$[x^2-4x+(x-1)][x^2-4x-(x-1)]=0$$
$$(x^2-3x-1)(x^2-5x+1)=0$$
$$(x^2-3x-1)=0 \Rightarrow x = \frac{3 \pm \sqrt{13} }{2} $$
$$(x^2-5x+1)=0 \Rightarrow x = \frac{5 \pm \sqrt{21} }{2} $$
$$(\frac{3 + \sqrt{13} }{2})^2 = \frac{22 + 6\sqrt{13} }{4}$$
$$(\frac{3 - \sqrt{13} }{2})^2 = \frac{22 - 6\sqrt{13} }{4}$$
$$(\frac{5 + \sqrt{21} }{2})^2 = \frac{46 + 10\sqrt{21} }{4}$$
$$(\frac{5 - \sqrt{21} }{2})^2 = \frac{46 - 10\sqrt{21} }{4}$$
$$sum = \frac{22+22+46+46}{4} = 34$$
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