Rewrite the given equation as $$2x^2 +2x-4 - 2x\sqrt{2x^2-3}=0\implies \left(\sqrt{2x^2-3}\right)^2 - 2x\sqrt{2x^2-3} + x^2 -x^2 + 2x-1$$

Therefore, we have $$ \left(\sqrt{2x^2-3} -x\right)^2 - (x-1)^2 = 0 \implies \left(\sqrt{2x^2-3} -x + x -1\right) \left(\sqrt{2x^2-3} -x - x + 1\right)=0 $$

Set each term equal $0$ yields only two roots: $\pm\sqrt{2}$.