Consider all lines which meet the graph of $$y=2x^4+7x^3+3x-5$$ in four distinct points, say $(x_i,y_i), i=1,2,3,4.$ Show that $$\frac{x_1+x_2+x_3+x_4}{4}$$ is independent of the line and find its value.
1. Consider the polynomial function \( f(x) = 2x^4 + 7x^3 + 3x - 5 \).
2. Let a line \( y = mx + b \) intersect the graph of \( f(x) \) at four distinct points \((x_i, y_i)\) for \( i = 1, 2, 3, 4 \).
3. At each intersection point, the coordinates \((x_i, y_i)\) satisfy both the equation of the line and the polynomial function. Therefore, we have:
   \[
   2x_i^4 + 7x_i^3 + 3x_i - 5 = mx_i + b
   \]
4. Rearrange the equation to form a polynomial equation:
   \[
   2x_i^4 + 7x_i^3 + (3 - m)x_i - (b + 5) = 0
   \]
5. This is a polynomial of degree 4 in \( x \). According to Vieta's formulas, the sum of the roots \( x_1, x_2, x_3, x_4 \) of the polynomial \( 2x^4 + 7x^3 + (3 - m)x - (b + 5) = 0 \) is given by the coefficient of \( x^3 \) divided by the leading coefficient (with a sign change):
   \[
   x_1 + x_2 + x_3 + x_4 = -\frac{\text{coefficient of } x^3}{\text{leading coefficient}} = -\frac{7}{2}
   \]
6. The arithmetic mean of the roots \( x_1, x_2, x_3, x_4 \) is:
   \[
   \frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{-\frac{7}{2}}{4} = -\frac{7}{8}
   \]
7. This value is independent of the line \( y = mx + b \) because it only depends on the coefficients of the polynomial \( 2x^4 + 7x^3 + (3 - m)x - (b + 5) \).

The final answer is \(\boxed{-\frac{7}{8}}\)