reserve x, a, b, c for Real;

theorem Th6:
  a > 0 & a * x^2 + b * x + c >= 0 implies (2 * a * x + b)^2 -
  delta(a,b,c) >= 0
proof
  assume that
A1: a > 0 and
A2: a * x^2 + b * x + c >= 0;
A3: 4 * a <> 0 by A1;
  4 * a > 0 & a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) >= 0 by A1,A2,Th1,
XREAL_1:129;
  then (4 * a) * (a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)) >= 0 by
XREAL_1:127;
  then
A4: ((2 * a) * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a
  )) >= 0;
  2 * a <> 0 by A1;
  then (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) >= 0 by A4,
XCMPLX_1:87;
  hence thesis by A3,XCMPLX_1:87;
end;
