We need to evaluate the following summation:

$\sum_{n=2}^{5} (-2n^2 + 3n)$

This means we will substitute $n = 2, 3, 4, 5$ into the expression $-2n^2 + 3n$ and sum the results.

Step-by-step evaluation:

1. For $n = 2$: $-2(2)^2 + 3(2) = -2(4) + 6 = -8 + 6 = -2$

2. For $n = 3$: $-2(3)^2 + 3(3) = -2(9) + 9 = -18 + 9 = -9$

3. For $n = 4$: $-2(4)^2 + 3(4) = -2(16) + 12 = -32 + 12 = -20$

4. For $n = 5$: $-2(5)^2 + 3(5) = -2(25) + 15 = -50 + 15 = -35$

Summing the results:

$-2 + (-9) + (-20) + (-35) = -66$

Thus, the value of the summation is:

$\boxed{-66}$

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. What happens if the summation range changes from $n = 1$ to $n = 5$?
2. How would the summation change if the expression inside was $2n^2 - 3n$ instead?
3. What are some common techniques for simplifying summations?
4. How does this summation relate to the concept of sequences in algebra?
5. How can you compute summations using sigma notation for higher powers of $n$?

Tip: In summations, it's often useful to check patterns in small cases to ensure your calculations are accurate.