Use algebra to find the set of values of x for which mod(3x^2 - 19x + 20) < 2x + 2.

The initial quadratic can be either positive or negative so we must solve for both possibilities.

Solving for positive:

3x^2 - 19x + 20 < 2x + 2    =    3x^2 - 21x + 18 < 0

                                           =    x^2 - 7x + 6 < 0

                                           =    (x - 6)(x - 1) < 0

Therefore, Critical Values where x crosses the x-axis are x = 1 or x = 6. And since we are solving for < 0, we focus on the graph under the x-axis, resulting in 1 < x < 6.

Solving for negative:

-3x^2 + 19x - 20 < 2x + 2    =    3x^2 - 17x + 22 > 0

                                            =    (3x - 11)(x - 2) > 0

Therefore, Critical Values where x crosses the x-axis are x = 2 or x = 11/3. And since we are solving for > 0, we focus on the graph above the x-axis, resulting in x < 2 or x > 11/3.

We must then find the values for x which satisfies both positive and negative, and using a simple numberline we find 1 < x < 2 or 11/3 < x < 6 as our final answers.