Solve the following quations by using qardratic formula:
`(1)/(x-2)+(1)/(x-3)+(1)/(x-4)=0`

Answer

To solve the equation \(\frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} = 0\) using the quadratic formula, we will follow these steps:

### Step 1: Combine the fractions
We start by combining the fractions on the left side of the equation. We can rewrite the equation as:
\[ \frac{(x-3)(x-4) + (x-2)(x-4) + (x-2)(x-3)}{(x-2)(x-3)(x-4)} = 0 \]
This means we need to find a common denominator, which is \((x-2)(x-3)(x-4)\).

### Step 2: Expand the numerators
Now we will expand each term in the numerator:
1. \((x-3)(x-4) = x^2 - 7x + 12\)
2. \((x-2)(x-4) = x^2 - 6x + 8\)
3. \((x-2)(x-3) = x^2 - 5x + 6\)

Adding these together:
\[ (x^2 - 7x + 12) + (x^2 - 6x + 8) + (x^2 - 5x + 6) = 3x^2 - 18x + 26 \]

### Step 3: Set the numerator equal to zero
Since the whole fraction equals zero, we set the numerator equal to zero:
\[ 3x^2 - 18x + 26 = 0 \]

### Step 4: Apply the quadratic formula
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
where \(D = b^2 - 4ac\). Here, \(a = 3\), \(b = -18\), and \(c = 26\).

### Step 5: Calculate the discriminant
Calculate \(D\):
\[ D = (-18)^2 - 4 \cdot 3 \cdot 26 = 324 - 312 = 12 \]

### Step 6: Substitute values into the quadratic formula
Now substitute \(D\) back into the quadratic formula:
\[ x = \frac{-(-18) \pm \sqrt{12}}{2 \cdot 3} = \frac{18 \pm 2\sqrt{3}}{6} \]

### Step 7: Simplify the expression
This simplifies to:
\[ x = 3 \pm \frac{\sqrt{3}}{3} \]

### Final Solution
Thus, the solutions are:
\[ x = 3 + \frac{\sqrt{3}}{3} \quad \text{and} \quad x = 3 - \frac{\sqrt{3}}{3} \]