Prove that 3(x+1)(x+7)-(2x+5)^2 is never positive?

If 2-2=0;2-2=0 again;
then -(0) is never positive.
cf:--(x-2)^=y ;
cf: -i^2 = -1.

3(x + 1)(x + 7) - (2x + 5)^2
= 3x^2 + 24x + 21 - 4x^2 - 20x - 25
= -x^2 + 4x - 4
= -(x - 2)^2

y = -(x - 2)^2

If x - 2 is imaginary (i.e. any real number times i where i^2 = -1), then 3(x+1)(x+7)-(2x+5)^2 is positive.
For example, if x = 2 + i, then 3(x+1)(x+7)-(2x+5)^2 = 1