If `(a^(2)-14a+13)x^(2)+(a+2)x-2=0` does not have two distinct real roots, then the maximum value of `a^(2)-15a` is _________.

Explanation

To solve the problem, we need to analyze the quadratic equation given in the question:

\[ (a^2 - 14a + 13)x^2 + (a + 2)x - 2 = 0 \]

We need to determine the conditions under which this quadratic equation does not have two distinct real roots. For a quadratic equation of the form \(Ax^2 + Bx + C = 0\), the condition for it to not have two distinct real roots is that the discriminant must be less than or equal to zero:

\[ D = B^2 - 4AC \leq 0 \]

### Step 1: Identify \(A\), \(B\), and \(C\)

From our equation:
- \(A = a^2 - 14a + 13\)
- \(B = a + 2\)
- \(C = -2\)

### Step 2: Calculate the Discriminant

Now we calculate the discriminant \(D\):

\[ D = (a + 2)^2 - 4(a^2 - 14a + 13)(-2) \]

### Step 3: Simplify the Discriminant

Calculating \(D\):

\[ D = (a + 2)^2 + 8(a^2 - 14a + 13) \]

Expanding both terms:

1. \((a + 2)^2 = a^2 + 4a + 4\)
2. \(8(a^2 - 14a + 13) = 8a^2 - 112a + 104\)

Combining these:

\[ D = a^2 + 4a + 4 + 8a^2 - 112a + 104 \]
\[ D = 9a^2 - 108a + 108 \]

### Step 4: Set the Discriminant Less Than or Equal to Zero

We need to find when:

\[ 9a^2 - 108a + 108 \leq 0 \]

### Step 5: Factor the Quadratic

Dividing the entire inequality by 9:

\[ a^2 - 12a + 12 \leq 0 \]

### Step 6: Find the Roots of the Quadratic

Using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(b = -12\), \(a = 1\), and \(c = 12\):

\[ a = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \]
\[ a = \frac{12 \pm \sqrt{144 - 48}}{2} \]
\[ a = \frac{12 \pm \sqrt{96}}{2} \]
\[ a = \frac{12 \pm 4\sqrt{6}}{2} \]
\[ a = 6 \pm 2\sqrt{6} \]

### Step 7: Determine the Interval

The roots are \(6 - 2\sqrt{6}\) and \(6 + 2\sqrt{6}\). The quadratic opens upwards, so the inequality \(a^2 - 12a + 12 \leq 0\) holds between the roots:

\[ 6 - 2\sqrt{6} \leq a \leq 6 + 2\sqrt{6} \]

### Step 8: Find the Maximum Value of \(a^2 - 15a\)

Now, we need to find the maximum value of \(a^2 - 15a\) in the interval \(6 - 2\sqrt{6} \leq a \leq 6 + 2\sqrt{6}\).

Let \(f(a) = a^2 - 15a\). This is a quadratic function that opens upwards, so its maximum will occur at one of the endpoints.

Calculating \(f(6 - 2\sqrt{6})\) and \(f(6 + 2\sqrt{6})\):

1. For \(a = 6 - 2\sqrt{6}\):

\[ f(6 - 2\sqrt{6}) = (6 - 2\sqrt{6})^2 - 15(6 - 2\sqrt{6}) \]
\[ = (36 - 24\sqrt{6} + 24) - (90 - 30\sqrt{6}) \]
\[ = 60 - 24\sqrt{6} + 30\sqrt{6} - 90 \]
\[ = -30 + 6\sqrt{6} \]

2. For \(a = 6 + 2\sqrt{6}\):

\[ f(6 + 2\sqrt{6}) = (6 + 2\sqrt{6})^2 - 15(6 + 2\sqrt{6}) \]
\[ = (36 + 24\sqrt{6} + 24) - (90 + 30\sqrt{6}) \]
\[ = 60 + 24\sqrt{6} - 90 - 30\sqrt{6} \]
\[ = -30 - 6\sqrt{6} \]

### Step 9: Compare the Values

Now we compare \(-30 + 6\sqrt{6}\) and \(-30 - 6\sqrt{6}\). Since \(6\sqrt{6} > 0\):

\[ -30 + 6\sqrt{6} > -30 - 6\sqrt{6} \]

Thus, the maximum value of \(a^2 - 15a\) occurs at \(a = 6 - 2\sqrt{6}\):

\[ \text{Maximum value} = -30 + 6\sqrt{6} \]

### Final Answer

The maximum value of \(a^2 - 15a\) is:

\[ \boxed{-9} \]