We need to evaluate the sum $\sum_{i=4}^{20} (2i - 5)$.

1. Problem Description

We need to evaluate the sum $\sum_{i=4}^{20} (2i - 5)$.

2. Solution Steps

The summation can be broken into two separate summations:

$\sum_{i=4}^{20} (2i - 5) = \sum_{i=4}^{20} 2i - \sum_{i=4}^{20} 5$

First, consider $\sum_{i=4}^{20} 2i = 2\sum_{i=4}^{20} i$.

We can rewrite $\sum_{i=4}^{20} i$ as $\sum_{i=1}^{20} i - \sum_{i=1}^{3} i$.

We know that $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$.

Therefore, $\sum_{i=1}^{20} i = \frac{20(20+1)}{2} = \frac{20(21)}{2} = 10(21) = 210$

And, $\sum_{i=1}^{3} i = \frac{3(3+1)}{2} = \frac{3(4)}{2} = 6$.

So, $\sum_{i=4}^{20} i = 210 - 6 = 204$.

Then, $2\sum_{i=4}^{20} i = 2(204) = 408$.

Next, consider $\sum_{i=4}^{20} 5$.

Since $5$ is a constant, we are adding $5$ a total of $20 - 4 + 1 = 17$ times.

Therefore, $\sum_{i=4}^{20} 5 = 5(17) = 85$.

Finally, $\sum_{i=4}^{20} (2i - 5) = 408 - 85 = 323$.

3. Final Answer

323