Inequality Involving Epsilon (precalculus)

Solve the inequality and write answer in interval notation.

Let E = epsilon

|3(x + 2)^2 - 3x^2| < E, where E > 0

3(x + 2)^2 - 3x^2
=3x^2+12x+12-3x^2
=12x+12

|12x+12| < E
|12| | x+1 | < E
|x+1| < E/12
-E/12 < x+1 < E/12
-1 -E/12 < x < -1 + E/12

|3(x + 2)^2 - 3x^2| < E
(3(x + 2)^2 - 3x^2)^2 < E^2
(12x + 12)^2 < E^2
144x^2 + 288x + 144 - E^2 < 0

consider 144x^2 + 288x + 144 - E^2 = 0
then x = -1 +/- sqrt(288^2 - 576(144 - E^2))/288
x = -1 +/- E/12
x = -1 + E/12 or x = -1- E/12

so -1 - (E/12) < x < -1 + (E/12)

3(x + 2)^2 - 3x^2= 12x+12 so |12x+12| -(E+12)/12 < x < (E-12)/12