Solve each of the following quadratic equations:
`x^(2)+6x-(a^(2)+2a-8)=0`

Explanation

To solve the quadratic equation \( x^2 + 6x - (a^2 + 2a - 8) = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
Start by rewriting the equation in standard form:
\[ x^2 + 6x - (a^2 + 2a - 8) = 0 \]
This can be simplified to:
\[ x^2 + 6x + (8 - a^2 - 2a) = 0 \]

### Step 2: Identify coefficients
In the standard form \( ax^2 + bx + c = 0 \), we identify:
- \( a = 1 \)
- \( b = 6 \)
- \( c = 8 - a^2 - 2a \)

### Step 3: Factor the quadratic
To factor the quadratic, we need two numbers that multiply to \( c \) and add up to \( b \). We need to find \( p \) and \( q \) such that:
\[ p + q = 6 \quad \text{and} \quad p \cdot q = 8 - a^2 - 2a \]

### Step 4: Solve for roots using the quadratic formula
If factoring is complex, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (8 - a^2 - 2a)}}{2 \cdot 1} \]
\[ x = \frac{-6 \pm \sqrt{36 - 32 + 4a^2 + 8a}}{2} \]
\[ x = \frac{-6 \pm \sqrt{4a^2 + 8a + 4}}{2} \]
\[ x = \frac{-6 \pm 2\sqrt{a^2 + 2a + 1}}{2} \]
\[ x = -3 \pm \sqrt{(a + 1)^2} \]

### Step 5: Simplify the roots
The square root simplifies to:
\[ x = -3 \pm (a + 1) \]
This gives us two possible roots:
1. \( x = -3 + (a + 1) = a - 2 \)
2. \( x = -3 - (a + 1) = -a - 4 \)

### Final Roots
Thus, the roots of the equation \( x^2 + 6x - (a^2 + 2a - 8) = 0 \) are:
\[ x = a - 2 \quad \text{and} \quad x = -a - 4 \]