The equation given in the image is:

$|x^2 + 2x - 4| = 4$

To solve this, we need to consider two cases since it involves an absolute value:

Case 1: $x^2 + 2x - 4 = 4$

First, solve the equation without the absolute value:

$x^2 + 2x - 4 = 4$

Simplify:

$x^2 + 2x - 4 - 4 = 0$ $x^2 + 2x - 8 = 0$

Now, solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = 1$, $b = 2$, and $c = -8$:

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)}$ $x = \frac{-2 \pm \sqrt{4 + 32}}{2}$ $x = \frac{-2 \pm \sqrt{36}}{2}$ $x = \frac{-2 \pm 6}{2}$

Thus, the two solutions are:

$x = \frac{-2 + 6}{2} = 2 \quad \text{or} \quad x = \frac{-2 - 6}{2} = -4$

Case 2: $x^2 + 2x - 4 = -4$

Now, solve the equation for the negative case:

$x^2 + 2x - 4 = -4$

Simplify:

$x^2 + 2x - 4 + 4 = 0$ $x^2 + 2x = 0$

Factor the equation:

$x(x + 2) = 0$

Thus, the solutions are:

$x = 0 \quad \text{or} \quad x = -2$

Final Solution:

The solutions from both cases are:

$x = 2, -4, 0, -2$

Therefore, the complete solution set is:

$x = \{ -4, -2, 0, 2 \}$

Would you like further explanation on any part of this? Here are some related questions you might find helpful:

1. How do we verify solutions for absolute value equations?
2. Can you explain the quadratic formula in more detail?
3. What are other methods to solve quadratic equations, aside from the quadratic formula?
4. How does factoring help in solving quadratic equations?
5. Can absolute value equations have no solutions?

Tip: Always check both cases in absolute value equations, as they can yield multiple solutions.