Solve the following quations by using qardratic formula:
`abx^(2)+(b^(2)-ac)x-bc=0`

Answer

To solve the quadratic equation \( abx^2 + (b^2 - ac)x - bc = 0 \) using the quadratic formula, we will follow these steps:

### Step 1: Identify coefficients
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). In our equation:
- \( a = ab \)
- \( b = b^2 - ac \)
- \( c = -bc \)

### Step 2: Calculate the discriminant
The discriminant \( D \) is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (b^2 - ac)^2 - 4(ab)(-bc) \]
Simplifying this:
\[ D = (b^2 - ac)^2 + 4abc \]
Expanding \( (b^2 - ac)^2 \):
\[ D = b^4 - 2b^2ac + a^2c^2 + 4abc \]
Combining like terms:
\[ D = b^4 + a^2c^2 + 2b^2ac \]
This can be factored as:
\[ D = (b^2 + ac)^2 \]

### Step 3: Apply the quadratic formula
The quadratic formula for the roots of the equation is:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values of \( b \) and \( D \):
\[ x = \frac{-(b^2 - ac) \pm \sqrt{(b^2 + ac)^2}}{2(ab)} \]

### Step 4: Simplify the expression
Since \( \sqrt{(b^2 + ac)^2} = b^2 + ac \):
\[ x = \frac{-(b^2 - ac) \pm (b^2 + ac)}{2ab} \]
This gives us two cases for \( x \):
1. For the positive case:
\[ x_1 = \frac{-b^2 + ac + b^2 + ac}{2ab} = \frac{2ac}{2ab} = \frac{c}{b} \]
2. For the negative case:
\[ x_2 = \frac{-b^2 + ac - (b^2 + ac)}{2ab} = \frac{-2b^2}{2ab} = -\frac{b}{a} \]

### Final Solution
Thus, the solutions to the equation \( abx^2 + (b^2 - ac)x - bc = 0 \) are:
\[ x_1 = \frac{c}{b} \quad \text{and} \quad x_2 = -\frac{b}{a} \]