## Solving the Equation (x+3)^4 + 2(x+3)^2 - 8 = 0

This equation might look intimidating at first, but we can solve it using a clever substitution and the quadratic formula. Here's how:

### 1. Substitution

Let's simplify the equation by substituting a new variable. Let **y = (x + 3)^2**. This allows us to rewrite the equation as:

**y^2 + 2y - 8 = 0**

### 2. Quadratic Formula

Now we have a simple quadratic equation. We can solve for *y* using the quadratic formula:

**y = (-b ± √(b^2 - 4ac)) / 2a**

Where *a* = 1, *b* = 2, and *c* = -8. Plugging in the values, we get:

**y = (-2 ± √(2^2 - 4 * 1 * -8)) / 2 * 1**

**y = (-2 ± √(36)) / 2**

**y = (-2 ± 6) / 2**

This gives us two possible solutions for *y*:

**y1 = 2**

**y2 = -4**

### 3. Solving for x

Now we need to substitute back to find the values of *x*. Remember that **y = (x + 3)^2**.

**For y1 = 2:**

**(x + 3)^2 = 2**

- Take the square root of both sides: x + 3 = ±√2
- Isolate x: x = -3 ± √2

**For y2 = -4:**

**(x + 3)^2 = -4**

- This equation has no real solutions because the square of a real number cannot be negative.

### 4. Solution

Therefore, the solutions to the original equation (x+3)^4 + 2(x+3)^2 - 8 = 0 are:

**x = -3 + √2**

**x = -3 - √2**