Solve the following equations :
(i) `2x^(2)-(3+7i)x-3+9i=0`
(ii) `(2+i)x^(2)-(5-i)x+2(1-i)=0`

Explanation

To solve the given equations, we will use the quadratic formula, which is:

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

where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).

### Part (i) \( 2x^2 - (3 + 7i)x - (3 - 9i) = 0 \)

1. **Identify coefficients**:
- \( a = 2 \)
- \( b = -(3 + 7i) = -3 - 7i \)
- \( c = -3 + 9i \)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac \]
\[ D = (-3 - 7i)^2 - 4(2)(-3 + 9i) \]
\[ D = (9 + 42i - 49) + 24 - 72i \]
\[ D = -40 + (42 - 72)i = -40 - 30i \]

3. **Calculate \( \sqrt{D} \)**:
Let \( \sqrt{D} = a + bi \). Then:
\[ a^2 - b^2 = -40 \]
\[ 2ab = -30 \]
From \( 2ab = -30 \), we have \( ab = -15 \).

Solving these equations:
- From \( a^2 - b^2 = -40 \) and \( ab = -15 \), we can substitute \( b = -\frac{15}{a} \) into the first equation and solve for \( a \) and \( b \).

4. **Substitute back into the quadratic formula**:
\[ x = \frac{-(-3 - 7i) \pm \sqrt{D}}{2(2)} \]
\[ x = \frac{3 + 7i \pm \sqrt{-40 - 30i}}{4} \]

### Part (ii) \( (2 + i)x^2 - (5 - i)x + 2(1 - i) = 0 \)

1. **Identify coefficients**:
- \( a = 2 + i \)
- \( b = -(5 - i) = -5 + i \)
- \( c = 2(1 - i) = 2 - 2i \)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac \]
\[ D = (-5 + i)^2 - 4(2 + i)(2 - 2i) \]
\[ D = (25 - 10i - 1) - 4(4 + 2i - 4i) \]
\[ D = 24 - 10i - 4(4 - 2i) \]
\[ D = 24 - 10i - 16 + 8i = 8 - 2i \]

3. **Calculate \( \sqrt{D} \)**:
Let \( \sqrt{D} = a + bi \). Then:
\[ a^2 - b^2 = 8 \]
\[ 2ab = -2 \]

4. **Substitute back into the quadratic formula**:
\[ x = \frac{-(-5 + i) \pm \sqrt{D}}{2(2 + i)} \]
\[ x = \frac{5 - i \pm \sqrt{8 - 2i}}{4 + 2i} \]