The Algorithm Design Manual - Working with the Big Oh
It is difficult to avoid what is repeating what is in the text, but you have exactly:
$T(n) = (n-1)+(n-2)+\ldots+1+0 =n\tfrac{n-1}{2} = \tfrac12n^2 -\tfrac12n$
and as an upper bound taking each value up to $n-1$:
$T(n) \le (n-1)+(n-1)+\ldots+(n-1)+(n-1) =n(n-1) = n^2-n$
and as an lower bound taking values of at least $\tfrac n2$ down to $\tfrac n2$ and other down to $0$ $($assuming $n$ is even; the odd case gives $\tfrac14n^2+\tfrac14 n)$:
$T(n) \ge \tfrac n2 +\ldots +\tfrac n2 +0+\ldots +0 = \tfrac n2 \times \tfrac n2 +\tfrac n2 \times 0 = \tfrac14n^2$
Since all three are essentially of the order of $n^2$ up to a constant, you can conclude $T(n)$ is $O(n^2)$ and $\Omega(n^2)$ and so $\Theta(n^2)$