Let $$f(x) = 2x^2 - x - 1$$ and $$S = \{n \in \mathbb{Z} : |f(n)| \leq 800\}$$. Then, the value of $$\sum_{n \in S} f(n)$$ is equal to

The set $$S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$$.

The first condition $$|z - 3| \leq 1$$ describes a closed disk centered at $$(3, 0)$$ with radius 1.

For the second condition, let $$z = x + iy$$, so $$\bar{z} = x - iy$$:

$$z(4+3i) + \bar{z}(4-3i) = (x+iy)(4+3i) + (x-iy)(4-3i)$$

$$= (4x - 3y) + i(3x + 4y) + (4x - 3y) - i(3x + 4y) = 2(4x - 3y) = 8x - 6y$$

So the second condition is $$8x - 6y \leq 24$$, i.e., $$4x - 3y \leq 12$$.

We need the point in $$S$$ closest to $$4i$$, which is the point $$(0, 4)$$ in the Cartesian plane.

The line from $$(0,4)$$ to the center of the disk $$(3,0)$$ has direction $$(3, -4)$$ and the distance from $$(0,4)$$ to $$(3,0)$$ is $$\sqrt{9+16} = 5$$.

The closest point on the disk to $$(0,4)$$ lies along this line, at distance $$5 - 1 = 4$$ from $$(0,4)$$. This point is:

$$(0,4) + \dfrac{4}{5}(3,-4) = \left(\dfrac{12}{5}, 4 - \dfrac{16}{5}\right) = \left(\dfrac{12}{5}, \dfrac{4}{5}\right)$$

We need to check that this point satisfies the linear constraint: $$4 \cdot \dfrac{12}{5} - 3 \cdot \dfrac{4}{5} = \dfrac{48 - 12}{5} = \dfrac{36}{5} = 7.2 \leq 12$$. Yes, it does.

So $$\alpha = \dfrac{12}{5}$$ and $$\beta = \dfrac{4}{5}$$.

$$25(\alpha + \beta) = 25\left(\dfrac{12}{5} + \dfrac{4}{5}\right) = 25 \cdot \dfrac{16}{5} = 80$$

The answer is $$80$$.