Solve this for me

Question

(1.) (x+2)²≥2(x²+7)

Answer 1

(x+2)²≥2(x²+7),

x²+4x+4≥2x²+14,

0≥x²-4x+10,

x²-4x+10≤0,

(x²-4x+4)-4+10≤0,

(x-2)²≤-6, has no real solutions, because a square cannot be negative.

There are complex solutions (containing imaginary numbers. For example, x=2±i√6).