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Let us consider the quadratic equation: $$(m-1)x^2-2mx+2=0.$$ This equation has no real solutions if and only if its discriminant is negative.

The discriminant $D$ is: $$D=[-2m{]}^2-4(m-1)\cdot 2=4m^2-8(m-1)=4m^2-8m+8.$$ We require: $$4m^2-8m+8<0.$$ Divide both sides by $4$: $$m^2-2m+2<0.$$ But the quadratic $m^2-2m+2$ has discriminant: $$(-2{)}^2-4\cdot 1\cdot 2=4-8=-4<0.$$ So $m^2-2m+2$ is always positive for all real $m$.

Therefore, there is no real value of $m$ for which the equation has no real solutions.