Solve the equation:
`x^(2)-(5+i)x+(18-i)=0`

Answer

To solve the equation \( x^2 - (5+i)x + (18-i) = 0 \), we will use the quadratic formula:

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

### Step 1: Identify coefficients
In our equation, we have:
- \( a = 1 \)
- \( b = -(5+i) \)
- \( c = 18-i \)

### Step 2: Calculate \( b^2 - 4ac \)
First, we need to calculate \( b^2 \):
\[ b^2 = (-(5+i))^2 = (5+i)^2 = 25 + 10i - 1 = 24 + 10i \]

Next, we calculate \( 4ac \):
\[ 4ac = 4 \cdot 1 \cdot (18-i) = 72 - 4i \]

Now, we can find \( b^2 - 4ac \):
\[ b^2 - 4ac = (24 + 10i) - (72 - 4i) = 24 + 10i - 72 + 4i = -48 + 14i \]

### Step 3: Calculate the square root of \( b^2 - 4ac \)
We need to find \( \sqrt{-48 + 14i} \). Let \( z = a + bi \) such that:
\[ z^2 = -48 + 14i \]
This gives us:
\[ (a^2 - b^2) + 2abi = -48 + 14i \]

From this, we can equate the real and imaginary parts:
1. \( a^2 - b^2 = -48 \)
2. \( 2ab = 14 \)

From the second equation, we can express \( b \) in terms of \( a \):
\[ b = \frac{7}{a} \]

### Step 4: Substitute \( b \) into the first equation
Substituting \( b \) into the first equation:
\[ a^2 - \left(\frac{7}{a}\right)^2 = -48 \]
This simplifies to:
\[ a^2 - \frac{49}{a^2} = -48 \]
Multiplying through by \( a^2 \):
\[ a^4 + 48a^2 - 49 = 0 \]

### Step 5: Let \( u = a^2 \)
Let \( u = a^2 \), then we have:
\[ u^2 + 48u - 49 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:
\[ u = \frac{-48 \pm \sqrt{48^2 + 4 \cdot 49}}{2} \]
Calculating the discriminant:
\[ 48^2 + 196 = 2304 + 196 = 2500 \]
Thus:
\[ u = \frac{-48 \pm 50}{2} \]
Calculating the two possible values:
1. \( u = 1 \) (valid)
2. \( u = -49 \) (not valid since \( u = a^2 \) cannot be negative)

So, \( a^2 = 1 \), hence \( a = \pm 1 \).

### Step 7: Find \( b \)
Using \( b = \frac{7}{a} \):
- If \( a = 1 \), then \( b = 7 \).
- If \( a = -1 \), then \( b = -7 \).

Thus, the two possible values for \( z \) are:
1. \( z = 1 + 7i \)
2. \( z = -1 - 7i \)

### Step 8: Substitute back into the quadratic formula
Now substituting back into the quadratic formula:
\[ x = \frac{5+i \pm (1 + 7i)}{2} \]
Calculating both cases:
1. \( x = \frac{6 + 8i}{2} = 3 + 4i \)
2. \( x = \frac{4 - 6i}{2} = 2 - 3i \)

### Final Solution
The solutions to the equation are:
\[ x = 3 + 4i \quad \text{and} \quad x = 2 - 3i \]