reserve x, a, b, c for Real;

theorem Th1:
  for a, b, c, x being Complex holds a <> 0 implies a * x^2
  + b * x + c = a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)
proof
  let a, b, c, x be Complex;
  assume
A1: a <> 0;
  then
A2: 4 * a <> 0;
  a * x^2 + b * x + c = a * x^2 + (b * x) * 1 + c
    .= a * x^2 + (b * x) * (a * (1/a)) + c by A1,XCMPLX_1:106
    .= a * (x^2 + ((b * x) * (1/a))) + c
    .= a * (x^2 + ((b * x)/a)) + c by XCMPLX_1:99
    .= a * (x^2 + x * (b/a)) + c by XCMPLX_1:74
    .= a * (x^2 + x * ((2 * b)/(2 * a))) + c by XCMPLX_1:91
    .= a * (x^2 + x * (2 * (b/(2 * a)))) + c by XCMPLX_1:74
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + b^2/(4 * a)) + (c - b^2/(4 * a))
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + a * (b^2/(4 * a) * (1/a))) + (c -
  b^2/(4 * a)) by A1,XCMPLX_1:109
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + a * ((b^2 * 1)/((4 * a) * a))) + (
  c - b^2/(4 * a)) by XCMPLX_1:76
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + a * (b^2/(2 * a)^2)) + (c - b^2/(4
  * a))
    .= (a * (x^2 + 2 * x * (b/(2 * a))) + a * (b/(2 * a))^2) + (c - b^2/(4 *
  a)) by XCMPLX_1:76
    .= a * (x + b/(2 * a))^2 - (b^2/(4 * a) - c)
    .= a * (x + b/(2 * a))^2 - (b^2/(4 * a) - ((4 * a * c)/(4 * a))) by A2,
XCMPLX_1:89
    .= a * (x + b/(2 * a))^2 - (b^2 - (4 * a * c))/(4 * a) by XCMPLX_1:120;
  hence thesis;
end;
