Maths again ... i'm stuck!?

Question

N=x^2-2x*[x^2+(sqrt5-x)*(sqrt5+x)]+25
we have to show that N => 0
means N it should be a positif number ... isn't it?

example of sqrt(i'm not from US or GB so i just want to make sure that you understood that!) sqrt9=3\sqrt81=9 :)

i saw sqrt on WinXP calculator :)

It is N=x^2-2x*[x^2+(sqrt5-x)*
*(sqrt5+x)]+25

Answer 1

N=x^2 - 2x*[x^2 + (√5-x)*(√5+x)] + 25
= x^2 - 2x[x^2 + (5 - √5 x + √5 x - x^2)] + 25
= x^2 - 2x(x^2 + 5 - x^2) + 25
= x^2 - 10x + 25
= (x-5)^2
≥ 0.