2014 AMC 10B Problem 20

First, note that $$x^4-51x^2+50$$$$= (x^2-50)(x^2-1).$$ The product is negative exactly when its two factors have opposite signs. Since $x^2-50<x^2-1$, this requires $x^2-50<0<x^2-1$. Thus $1<x^2<50$, or $2\le |x|\le7$. There are $6$ positive and $6$ negative integer solutions, for a total of $12$. Thus, the correct answer is C .