We need to split the problem into two cases to remove the abs sign. Case 1: 5x - 6 >= 0, equivalent to x >= 6/5. In this case the problem is now to find the values of x that satisfy x2 > 5x - 6. Rearranging and factorising gives (x-3)(x-2) > 0. By sketching the graphs to this, it's easy to see it is greater than zero when x < 2 and x > 3. In this case, we had the constraint that x >= 6/5. So the satisfying values for case 1 is 6/5 <= x < 2, and x > 3.
Case 2: 5x - 6 < 0, equivalent to x < 6/5. Now the problem is x2 > -(5x - 6), or after rearranging, x2 + 5x - 6 > 0. Factorising gives (x+6)(x-1) > 0. By sketching this graph, it's easy to see it is greater than zero when x < -6, and when x > 1. In this case, we had the constraint that x < 6/5. So the satisfying values for case 2 is x < -6, and 1 < x < 6/5. Bringing the two cases together we have x < -6, 1 < x < 6/5, 6/5 <= x < 2, and x > 3. This simplifies to x < -6, 1 < x < 2, x > 3.