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 follow these steps:

### Step 1: Identify coefficients
The given quadratic equation can be rewritten in the standard form \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = 1 \)
- \( b = 5 \)
- \( c = -(a^2 + a - 6) \)

### Step 2: Write the quadratic formula
The quadratic formula to find the roots of the equation \( ax^2 + bx + c = 0 \) is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 3: Substitute the values into the formula
Substituting the identified values into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-(a^2 + a - 6))}}{2 \cdot 1} \]

### Step 4: Simplify the expression under the square root
Calculating \( b^2 - 4ac \):
\[ b^2 = 5^2 = 25 \]
\[ 4ac = 4 \cdot 1 \cdot (-(a^2 + a - 6)) = -4(-a^2 - a + 6) = 4a^2 + 4a - 24 \]
Thus,
\[ b^2 - 4ac = 25 + (4a^2 + 4a - 24) = 4a^2 + 4a + 1 \]

### Step 5: Substitute back into the formula
Now substituting back into the formula:
\[ x = \frac{-5 \pm \sqrt{4a^2 + 4a + 1}}{2} \]

### Step 6: Simplify the square root
Notice that \( 4a^2 + 4a + 1 \) can be factored as:
\[ 4a^2 + 4a + 1 = (2a + 1)^2 \]
Thus, we have:
\[ x = \frac{-5 \pm (2a + 1)}{2} \]

### Step 7: Calculate the two possible values for \( x \)
Calculating the two possible values for \( x \):
1. First root:
\[ x_1 = \frac{-5 + (2a + 1)}{2} = \frac{2a - 4}{2} = a - 2 \]

2. Second root:
\[ x_2 = \frac{-5 - (2a + 1)}{2} = \frac{-2a - 6}{2} = -a - 3 \]

### Final Solution
The solutions for the quadratic equation \( x^2 + 5x - (a^2 + a - 6) = 0 \) are:
\[ x = a - 2 \quad \text{and} \quad x = -a - 3 \]