Binomial Theorem Question

Quote:

Prove that n has no solutions given that r=2 when:

$\displaystyle

n!/

r!(n-r)!

= 4

$

I expanded this to:

$\displaystyle

n(n-1)(n-2)(n-3)...1/

2((n-2)(n-3)(n-4)...1)

= 4

$

Which cancels down to:

$\displaystyle

n(n-1)/

2

= 4

$

Which goes to:

$\displaystyle

n^2-n

= 8

$

Which goes to:

$\displaystyle

n^2-n - 8

= 0

$

This is now a quadratic which, when treated with the discriminant (a=1, b=-1, c=-8)

$\displaystyle b^2-4ac$

$\displaystyle (-1)^2-4*1*-8$

$\displaystyle =1+32 = 33$

So, this quadratic has real solutions, but the questions asks to prove it **does not**. What have I done wrong?

Thanks.