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Given $$(m-2)x^2-4x+3=3-x^2+2nx$$ Compute the value of $m$.

$$(m-2)x^2-4x+3=3-x^2+2nx \\mx^2+(-4m-4)x+7=-x^2+2nx+3 \\m=-1, n=0$$

The answer is $m=1,n=-2$, please tell me where did I go wrong

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    $\begingroup$ There is no linear term in $(m-2)x^2$, it's purely quadratic in $x$. $\endgroup$ – lulu Sep 23 '18 at 13:38
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Hint:

The coefficient of $x^2$ on the left of the equal sign is $(m-2)$ and on the right it is $-1$

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