Ta có :
`(-2)^3-(m^2-m+7)(-2)-3(m^2-m-2)=0`
`<=>-8+2(m^2-m+7)(-2)-3(m^2-m-2)=0`
`<=>-m^2+m+12=0`
`<=>(m-4)(m+2)=0`
`<=>` \(\left[ \begin{array}{l}m=4\\m=-3\end{array} \right.\)
Nếu $m=4$ thì `x^3+19x-30=0`
`<=>` \begin{cases}x^3+2x^2-2n^2-4x-15x-30=0\\n^2(n+2)-2x(n+2)-15(x+2)=0\end{cases}
`<=>` \begin{cases}(x+2)(x^2-2n-1)=0\\(x+2).(n-5)(n+3)=0\end{cases}
`<=>` \(\left[ \begin{array}{l}x=-2\\x=5\\x=-3\end{array} \right.\)
Nếu `m=-3` thì : `x^3+19x-30=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=5\\x=-3\end{array} \right.\)
Vậy `x∈{-2;5;-3}`