[You forgot to tell us what f(x) is.  So I made up one that
 would work the same way, f(x) = x²+4x+1.

 To show that f(-2+2i) is a real number, we substitute (-2+2i)
 for x in

 f(x) = x²+4x+1

 f(-2+2i) = (-2+2i)²+4(-2+2i)+1

 f(-2+2i) = (-2+2i)(-2+2i)-8+8i+1

 f(-2+2i) = (4-4i-4i+4i²)-8+8i+1
 
 f(-2+2i) = 4-4i-4i+4i²-8+8i+1

 f(-2+2i) = 4-8i+4i²-8+8i+1

 Since i² = -1, we can substitute (-1) for i²

 f(-2+2i) = 4-8i+4(-1)-8+8i+1

 f(-2+2i) = 4-8i-4-8+8i+1

 The -8i and the 8i cancel out.

 f(-2+2i) = 4-4-8+1

 f(-2+2i) = -7

 And -7 is a real number because it does not have any i's,
 since the terms that had i's all canceled out.

 ---------------------

 To show that f(7+4i) is NOT a real number, we substitute (7+4i)
 for x in

 f(x) = x²+4x+1

 f(7+4i) = (7+4i)²+4(7+4i)+1

 f(7+4i) = (7+4i)(7+4i)+28+16i+1

 f(7+4i) = (49+28i+28i+16i²)+28+16i+1

 f(7+4i) = 49+28i+28i+16i²+28+16i+1

 f(7+4i) = 49+56i+16i²+28+16i+1

 Since i² = -1, we can substitute (-1) for i²

 f(7+4i) = 49+56i+16(-1)+28+16i+1

 f(7+4i) = 49+56i-16+28+16i+1

 f(7+4i) = 49-16+28+1+56i+16i

 f(7+4i) = 62+72i

 There is still an i in 62+72i, so that is NOT a real number.

 Now do the same thing with the equation for f(x) which you
 were given that you forgot to type above.

 Edwin