<s>[INST] Solve the equation $(m^2-m-2)x=m^2+4m+3$ Here's how I solve it
I think that m is the variable (am I right?).
Then
$$m^2x-mx-2x-m^2-4m-3=0$$
$$m^2(x-1)-m(x+4)-(2x+3)=0$$
$$D=x^2+8x+16+4(x-1)(2x+3)$$
 $$=x^2+8x+16+4(2x^2-2x+3x-3)$$
 $$=9x^2+12x+4$$
 $$=(3x+2)^2$$
$$m=\frac{x+4\pm (3x-2))}{2(x-1)}$$
$$m_1=\frac{4x+2}{2x-2}$$
$$m_2=\frac{-2x+6}{2x-2}$$
Is this right?
I don't know if the tag is right, so please don't be based on it.
 [/INST] $m^2-m-2=(m-2)(m+1)$ and $m^2+4m+3=(m+1)(m+3)$
Clearly $m+1=0$ is a solution
Else $(m-2)x=m+3\implies x=\dfrac{m+3}{m-2}=1+\dfrac5{m-2}$
If $x$ is an integer, $(m-2)$ must divide $5$
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