Determine Sum of $\sum_{k=3}^{25}{k^2-5k+9}$ Given $\sum_{k=3}^{28}(k-3)^2 = 14,910$ and $\sum_{k=0}^{25}k = 325$

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I am working through some questions from a textbook which states that I should determine the sum without expanding or calculating any sums. I have been given the following information.

$\sum_{k=3}^{28}(k-3)^2 = 14,910$ and $\sum_{k=0}^{25}k = 325$

And must calculate the sum for $\sum_{k=3}^{25}{k^2-5k+9}$

So far I've noticed the following,

${\sum}_{k=3}^{28}\left(k-3\right)^{2}={\sum}_{k=0}^{25}k^{2}$

${\sum}_{k=0}^{25}k={\sum}_{k=3}^{28}k-3$

I'm also noticed that, $\left(k-3\right)^{2}=k^{2}-6k+9$, is very close to the sum I need to find, but am not sure how this helps.

I calculated the sum as follows, but I think the textbook would not have wanted me to calculate it in this way,

$\sum_{k=3}^{25}{k^2-5k+9}={\sum}_{k=3}^{25}k^{2}-5{\sum}_{k=3}^{25}k+{\sum}_{k=3}^{25}9=14,910-5-5\times\left(325-3\right)+22\times9=13,493$

I believe my answer is wrong however I'm not sure how to proceed. I would appreciate it if someone could provide some sort of hint to guide me in the right direction.

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We can manually subtract terms from each of the two sums to get the desired range: $$\sum_{k=3}^{25}(k-3)^2=14910-25^2-24^2-23^2=13180$$ $$\sum_{k=3}^{25}k=325-0-1-2=322$$ The final answer is $13180+322=13502$.


In fact, the given values are completely wrong; the actual correct answer is $4117$.