reserve x, a, b, c for Real;

theorem Th21:
  a > 0 & (2 * a * x + b)^2 - delta(a,b,c) > 0 implies a * x^2 + b * x + c > 0
proof
  assume that
A1: a > 0 and
A2: (2 * a * x + b)^2 - delta(a,b,c) > 0;
  4 * a <> 0 by A1;
  then
A3: (2 * a * x + b)^2 - (4 * a) * (delta(a,b,c)/(4 * a)) > 0 by A2,XCMPLX_1:87;
  2 * a <> 0 by A1;
  then
  (2 * a * x + (2 * a) * (b/(2 * a)))^2 - (4 * a) * (delta(a,b,c)/(4 * a))
  > 0 by A3,XCMPLX_1:87;
  then
A4: (4 * a) * (a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a)) > 0;
  4 * a > 0 by A1,XREAL_1:129;
  then a * (x + b/(2 * a))^2 - delta(a,b,c)/(4 * a) > 0/(4 * a) by A4,
XREAL_1:83;
  hence thesis by A1,Th1;
end;
