Use summation rules to find the following: \[ \sum_{k=1}^{28}\left(2 k^{2}-6\right)= \]

Step 1 ：The given summation can be broken down into two separate summations: one for \(2k^2\) and one for \(-6\).

Step 2 ：The summation of \(2k^2\) from \(k=1\) to \(28\) can be found using the formula for the sum of squares, which is \(\frac{n(n+1)(2n+1)}{6}\). Substituting \(n=28\) into the formula, we get \(15428\).

Step 3 ：The summation of \(-6\) from \(k=1\) to \(28\) is simply \(-6\) times \(28\), which equals \(-168\).

Step 4 ：Adding these two results together, we get \(15428 - 168 = 15260\).

Step 5 ：Final Answer: The sum of the series \(\sum_{k=1}^{28}\left(2 k^{2}-6\right)\) is \(\boxed{15260}\).