2 Expressions make a Number

$If\quad \frac { f(x) }{ g(x) } is\quad a\quad natural\quad number,\quad g(x)=1\quad and\quad x\quad is\quad a\quad natural\quad number\quad OR\quad f(x)=kg(x)\quad where\quad k=natural\quad number\quad constant\quad and\quad x\quad is\quad a\quad real\quad number$ First case: $g(x)=1\rightarrow x^{ 2 }-x+2=1\rightarrow x^{ 2 }-x+1=0\rightarrow It\quad has\quad no\quad solution.$ Second case: $f(x)=kg(x)\rightarrow 2{ x }^{ 2 }+2x-4=k({ x }^{ 2 }-x+2)\rightarrow (k-2){ x }^{ 2 }-(k+2)x+(2k+4)=0.\\ In\quad order\quad for\quad the\quad last\quad equation\quad to\quad have\quad a\quad real\quad solution,\quad D\ge 0\\ D={ (k+2) }^{ 2 }-4(k-2)(2k+4)=-7{ k }^{ 2 }+4k+36\ge \rightarrow (k-2)(7k+18)\le 0\\ \therefore -\frac { 18 }{ 7 } \le k\le 2.\quad Since\quad k\quad is\quad a\quad natural\quad number,\quad k=1\quad or\quad 2\\ \\ If\quad k=1,\quad f(x)=g(x)\rightarrow 2{ x }^{ 2 }+2x-4={ x }^{ 2 }-x+2\rightarrow { x }^{ 2 }+3x-6=0\\ If\quad k=2,\quad f(x)=2g(x)\rightarrow 2{ x }^{ 2 }+2x-4=2({ x }^{ 2 }-x+2)\rightarrow x=2\\ \\ \therefore The\quad sum\quad of\quad all\quad the\quad x\quad values\quad is\quad -3+2=-1$