Solve this inequality. Verify your results using a graphing utility. x2 + x > 30

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James is correct.  Your related equation is (x + 6)(x - 5) = 0.  Thus your two x-intercepts are -6 and 5.  Since your leading coefficient (which is 1, the coefficient in the quadratic term) is positive, then your graph will open upwards.  Sketch a parabola opening upward and passing through the points (-6,0) and (5,0).  Looking back at the inequality, we have x^2 + x -30 > 0 or f(x) > 0.  Thus, we are looking for values of f(x) that are greater than zero (or above the x-axis).  The only part of our function's graph that is above the x-axis are the ends which stretch up and to the left before x = -6 and up and to the right after x = 5.  To continue graphing, draw open circles over the intercepts.  ***They'd be closed dots if it were greater than or equal to.  Then, draw arrows stretching outward from the intercepts to indicate that those are the values of x which make the inequality true.  Your solution is {x|x<-6 or x>5}.