reserve x,y,a,b,c,p,q for Real;
reserve m,n for Element of NAT;

theorem
  a<>0 & b^2-2*a*b-3*a^2>=0 & Polynom(a,b,b,a,x)=0 implies x=-1 or x= (a
  -b+sqrt(b^2-2*a*b-3*a^2))/(2*a) or x= (a-b-sqrt(b^2-2*a*b-3*a^2))/(2*a)
proof
  assume that
A1: a<>0 & b^2-2*a*b-3*a^2>=0 and
A2: Polynom(a,b,b,a,x)=0;
  a*(x|^ 3)+b*x^2+b*x+a=0 by A2,POLYEQ_1:def 4;
  then ((x|^ 3)+1)*a+(x^2+x+0)*b=0;
  then ((x|^ 3)+1 to_power 3)*a+(x+1)*x*b=0;
  then ((x+1)*(x^2-x*1+1^2))*a+(x+1)*x*b=0 by Th11;
  then
A3: (x^2*a-x*a+x*b+a)*(x+1)=0;
  now
    per cases by A3;
    case
      x+1=0;
      hence thesis;
    end;
    case
A4:   a*x^2-(a-b)*x+a=0;
A5:   delta(a,-a+b,a) =(-a+b)^2-4*a*a by QUIN_1:def 1
        .=(b^2-2*a*b)+(-(4-1))*a^2;
      a*x^2+(-a+b)*x+a=0 by A4;
      then Polynom(a,-a+b,a,x)=0 by POLYEQ_1:def 2;
      then
      x= (-(-a+b)+sqrt delta(a,-a+b,a))/(2*a) or x= (-(-a+b)-sqrt delta(a
      ,-a+b,a))/(2*a) by A1,A5,POLYEQ_1:5;
      hence thesis by A5;
    end;
  end;
  hence thesis;
end;
