Will squaring both sides introduce a $x^4$ term that may complicate the equation?

**Hint:**

$$sqrt{2x^2 – 4x + 3} = sqrt{2(x^2 – 2x) + 3}$$

Let $v = x^2 – 2x$.

$$implies x^2 – 2x = sqrt{2(x^2 – 2x) + 3}longrightarrow v = sqrt{2v + 3}$$

Square both sides to get

$$v^2 = 2v + 3 \ v^2 – 2v – 3 = 0$$

Solve the quadratic equation to find the roots of $v$ and then undo the substitution to find the values of $x$ when $v > 0$.

Squaring both sides we have $