We are tasked with solving the inequality:

$2|x + 2| - 8 > 14$

Step 1: Isolate the absolute value expression

First, add 8 to both sides to move the constant term on the left-hand side:

$2|x + 2| > 22$

Now, divide both sides by 2:

$|x + 2| > 11$

Step 2: Break into two cases

An absolute value inequality of the form $|A| > B$ breaks into two cases:

1. $x + 2 > 11$
2. $x + 2 < -11$

Case 1: $x + 2 > 11$

Solve for $x$:

$x > 11 - 2$ $x > 9$

Case 2: $x + 2 < -11$

Solve for $x$:

$x < -11 - 2$ $x < -13$

Step 3: Combine the two cases

The solution is:

$x > 9 \quad \text{or} \quad x < -13$

In interval notation, this is:

$(-\infty, -13) \cup (9, \infty)$

This is the solution to the inequality.

Would you like more details on any step or have any questions?

Related Questions:

1. How do we solve inequalities with absolute values when the sign is "less than" instead of "greater than"?
2. What happens if the constant on the right side of the absolute value inequality is negative?
3. Can you explain how to graph the solution of an absolute value inequality?
4. How would the solution change if there was a different coefficient before the absolute value term?
5. How do you solve an equation with absolute value and quadratic terms?

Tip:

When solving absolute value inequalities, always remember to split the problem into two cases—one for the positive scenario and one for the negative.