
theorem
  for a,c,e,x being Real st a <> 0 & e <> 0 & c^2 - (4*a*e) > 0
holds Polynom(a,0,c,0,e,x) = 0 implies x <> 0 & (x = sqrt((-c + sqrt delta(a,c,
e))/(2*a)) or x = sqrt((-c - sqrt delta(a,c,e))/(2*a)) or x = - sqrt((-c + sqrt
  delta(a,c,e))/(2*a)) or x = - sqrt((-c - sqrt delta(a,c,e))/(2*a)))
proof
  let a,c,e,x be Real;
  assume that
A1: a <> 0 and
A2: e <> 0 and
A3: c ^2 -(4*a*e) > 0;
  set y = x^2;
  assume
A4: Polynom(a,0,c,0,e,x) = 0;
A5: now
    assume x = 0;
    then a*0 +0*(0 |^ 3)+ e = 0 by A4,NEWTON:11;
    hence contradiction by A2;
  end;
  per cases by A5,XXREAL_0:1;
  suppose
A6: x > 0;
    x|^ 4 = x to_power (2+2) by POWER:41
      .= (x to_power 2)*(x to_power 2) by A6,POWER:27
      .= x^2*(x to_power 2) by POWER:46
      .= x^2*x^2 by POWER:46;
    then a*y^2+c*y + e = 0 by A4;
    then
A7: Polynom (a,c,e,y) = 0 by POLYEQ_1:def 2;
    delta(a,c,e) >= 0 by A3,QUIN_1:def 1;
    then
    y = (-c + sqrt delta(a,c,e))/(2*a) or y = (-c - sqrt delta(a,c,e))/(2
    *a) by A1,A7,POLYEQ_1:5;
    then |.x.| = sqrt((-c + sqrt delta(a,c,e))/(2*a)) or |.x.| = sqrt((-c -
    sqrt delta(a,c,e))/(2*a)) by COMPLEX1:72;
    hence thesis by A6,ABSVALUE:def 1;
  end;
  suppose
A8: x < 0;
    then
A9: 0 < - x by XREAL_1:58;
    (-x)|^ 4 = x|^ 4 by Lm1,POWER:1;
    then x|^ 4 = (-x) to_power (2+2) by POWER:41
      .= ((-x) to_power 2)*((-x) to_power 2) by A9,POWER:27
      .= (-x)^2*((-x) to_power 2) by POWER:46
      .= x^2*x^2 by POWER:46;
    then a*y^2+c*y + e = 0 by A4;
    then
A10: Polynom(a,c,e,y) = 0 by POLYEQ_1:def 2;
    (c^2-4*a*e) = delta(a,c,e) by QUIN_1:def 1;
    then
    y = (-c + sqrt delta(a,c,e))/(2*a) or y = (-c - sqrt delta(a,c,e))/(2
    *a) by A1,A3,A10,POLYEQ_1:5;
    then |.x.| = sqrt((-c + sqrt delta(a,c,e))/(2*a)) or |.x.| = sqrt((-c -
    sqrt delta(a,c,e))/(2*a)) by COMPLEX1:72;
    then
    (-1)*(-x) = (-1)* sqrt((-c + sqrt delta(a,c,e))/(2*a)) or (-1)*(-x) =
    (-1)*sqrt((-c - sqrt delta(a,c,e))/(2*a)) by A8,ABSVALUE:def 1;
    hence thesis by A5;
  end;
end;
