Solve the Equation:
`x^(2)-(2+i)x=1-7i`

Answer

To solve the equation \( x^2 - (2+i)x = 1 - 7i \), we will follow these steps:

### Step 1: Rearrange the equation
We start by rearranging the equation to bring all terms to one side:
\[ x^2 - (2+i)x - (1 - 7i) = 0 \]
This can be simplified to:
\[ x^2 - (2+i)x + (7i - 1) = 0 \]

### Step 2: Identify coefficients
Now, we identify the coefficients \( a \), \( b \), and \( c \) in the standard quadratic form \( ax^2 + bx + c = 0 \):
- \( a = 1 \)
- \( b = -(2+i) \)
- \( c = 7i - 1 \)

### Step 3: Apply the quadratic formula
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{2+i \pm \sqrt{(2+i)^2 - 4 \cdot 1 \cdot (7i - 1)}}{2} \]

### Step 4: Calculate \( b^2 \)
Now, we calculate \( b^2 \):
\[ (2+i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i \]

### Step 5: Calculate \( 4ac \)
Next, we calculate \( 4ac \):
\[ 4 \cdot 1 \cdot (7i - 1) = 4(7i - 1) = 28i - 4 \]

### Step 6: Calculate \( b^2 - 4ac \)
Now we find \( b^2 - 4ac \):
\[ (3 + 4i) - (28i - 4) = 3 + 4i - 28i + 4 = 7 - 24i \]

### Step 7: Find the square root of \( 7 - 24i \)
Let \( z = x + yi \) such that \( z^2 = 7 - 24i \). This gives us:
\[ x^2 - y^2 + 2xyi = 7 - 24i \]
From here, we equate real and imaginary parts:
1. \( x^2 - y^2 = 7 \)
2. \( 2xy = -24 \)

From the second equation, we can express \( y \) in terms of \( x \):
\[ y = \frac{-12}{x} \]

### Step 8: Substitute \( y \) into the first equation
Substituting \( y \) into the first equation:
\[ x^2 - \left(\frac{-12}{x}\right)^2 = 7 \]
This simplifies to:
\[ x^2 - \frac{144}{x^2} = 7 \]
Multiplying through by \( x^2 \) to eliminate the fraction:
\[ x^4 - 7x^2 - 144 = 0 \]

### Step 9: Let \( u = x^2 \)
Let \( u = x^2 \), so we have:
\[ u^2 - 7u - 144 = 0 \]
Using the quadratic formula:
\[ u = \frac{7 \pm \sqrt{49 + 576}}{2} = \frac{7 \pm 25}{2} \]
This gives us:
\[ u = 16 \quad \text{or} \quad u = -9 \]

### Step 10: Solve for \( x \)
Since \( u = x^2 \), we only take the positive solution:
\[ x^2 = 16 \implies x = \pm 4 \]
From \( y = \frac{-12}{x} \):
- If \( x = 4 \), then \( y = -3 \)
- If \( x = -4 \), then \( y = 3 \)

Thus, the solutions for \( z \) are:
\[ z = 4 - 3i \quad \text{and} \quad z = -4 + 3i \]

### Step 11: Substitute back to find \( x \)
Now substituting back to find \( x \):
\[ x = \frac{2+i \pm (4 - 3i)}{2} \quad \text{and} \quad x = \frac{2+i \pm (-4 + 3i)}{2} \]
Calculating these gives us:
1. \( x = \frac{6 - 2i}{2} = 3 - i \)
2. \( x = \frac{-2 + 4i}{2} = -1 + 2i \)

### Final Solutions
The final solutions are:
\[ x = 3 - i \quad \text{and} \quad x = -1 + 2i \]