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

This equation might look intimidating at first glance, but we can solve it using a strategic substitution and some algebraic manipulation. Here's how:

### 1. Simplify by Substitution

Let's substitute **y = x^2 + 3x**. This simplifies our equation to:

**y^2 - 8y - 4 = 0**

This is now a quadratic equation in terms of 'y'.

### 2. Solve the Quadratic Equation

We can solve for 'y' using the quadratic formula:

**y = (-b ± √(b^2 - 4ac)) / 2a**

Where a = 1, b = -8, and c = -4.

Plugging in the values:

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

**y = (8 ± √(80)) / 2**

**y = (8 ± 4√5) / 2**

**y = 4 ± 2√5**

We now have two possible solutions for 'y'.

### 3. Substitute Back and Solve for 'x'

Let's substitute back the original expression for 'y':

**Case 1: y = 4 + 2√5**

**(x^2 + 3x) = 4 + 2√5**

**x^2 + 3x - (4 + 2√5) = 0**

This is a quadratic equation in terms of 'x'. We can solve it using the quadratic formula again, but the solutions will involve radicals.

**Case 2: y = 4 - 2√5**

**(x^2 + 3x) = 4 - 2√5**

**x^2 + 3x - (4 - 2√5) = 0**

Again, we can solve for 'x' using the quadratic formula.

### 4. Finding the Solutions

The solutions for 'x' in both cases will be the roots of the quadratic equations obtained after substituting back the values of 'y'. These solutions will involve radicals.

**In summary:**

- We simplified the equation using substitution.
- Solved the resulting quadratic equation for 'y'.
- Substituted back the original expression for 'y' and solved for 'x' using the quadratic formula.

This process will give you the four possible solutions for 'x' in the original equation.