Solve `x^(2)+5x-(a^(2)+a-6)=0`.

Answer

To solve the quadratic equation \( x^2 + 5x - (a^2 + a - 6) = 0 \), we will use the quadratic formula. The quadratic formula states that for an equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by:

\[ x = \frac{-b \pm \sqrt{d}}{2a} \]

where \( d \) is the discriminant calculated as \( d = b^2 - 4ac \).

### Step 1: Identify coefficients
In our equation \( x^2 + 5x - (a^2 + a - 6) = 0 \), we can identify:
- \( a = 1 \) (coefficient of \( x^2 \))
- \( b = 5 \) (coefficient of \( x \))
- \( c = -(a^2 + a - 6) \)

### Step 2: Calculate the discriminant
Now, we calculate the discriminant \( d \):

\[ d = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-(a^2 + a - 6)) \]

Calculating this gives:

\[ d = 25 + 4(a^2 + a - 6) \]
\[ d = 25 + 4a^2 + 4a - 24 \]
\[ d = 4a^2 + 4a + 1 \]

### Step 3: Simplify the discriminant
Notice that \( 4a^2 + 4a + 1 \) can be factored:

\[ d = (2a + 1)^2 \]

### Step 4: Apply the quadratic formula
Now we can substitute \( d \) back into the quadratic formula:

\[ x = \frac{-5 \pm \sqrt{(2a + 1)^2}}{2 \cdot 1} \]

Since the square root of a square gives us the absolute value, we have:

\[ x = \frac{-5 \pm (2a + 1)}{2} \]

### Step 5: Calculate the two possible values of \( x \)
Calculating the two cases:

1. **Positive case**:
\[ x = \frac{-5 + (2a + 1)}{2} = \frac{2a - 4}{2} = a - 2 \]

2. **Negative case**:
\[ x = \frac{-5 - (2a + 1)}{2} = \frac{-6 - 2a}{2} = -3 - a \]

### Final Solution
Thus, the solutions for the equation \( x^2 + 5x - (a^2 + a - 6) = 0 \) are:

\[ x = a - 2 \quad \text{and} \quad x = -3 - a \]