If $\alpha, \beta$ are rational roots such that $\alpha < \beta$ and $\gamma, \delta$ are irrational roots such that $\gamma > \delta$ of the equation $(x+9)(x-3)(x-7)(x+5)=385$ then $\alpha- \beta- \gamma - \delta$ is

By regrouping the equation into (x^2 + 14x + 45)(x^2 - 10x - 21) = 385 and substituting m = x^2 + 2x, we get the quadratic (m + 45)(m - 21) = 385. Solving this gives m = 34 and m = -50, which leads to the two quadratics x^2 + 2x - 34 = 0 and x^2 + 2x + 50 = 0. The first quadratic provides the irrational roots gamma and delta as -1 + sqrt(35) and -1 - sqrt(35), while the second quadratic yields rational roots of 5 and -7. Calculating alpha - beta - gamma - delta with alpha = 5 and beta = -7 gives the result of 0.

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