Solve each of the following equatins :
`6x^(2)-x-2=0`

Answer

To solve the quadratic equation \( 6x^2 - x - 2 = 0 \) using the method of factorization, follow these steps:

### Step 1: Identify coefficients
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = 6 \)
- \( b = -1 \)
- \( c = -2 \)

### Step 2: Find two numbers that multiply to \( ac \) and add to \( b \)
We need to find two numbers that multiply to \( ac = 6 \times (-2) = -12 \) and add to \( b = -1 \). The two numbers that satisfy these conditions are \( -4 \) and \( 3 \) because:
- \( -4 + 3 = -1 \)
- \( -4 \times 3 = -12 \)

### Step 3: Rewrite the middle term using the two numbers
We can rewrite the equation by breaking the middle term:
\[ 6x^2 - 4x + 3x - 2 = 0 \]

### Step 4: Factor by grouping
Now, we group the terms:
\[ (6x^2 - 4x) + (3x - 2) = 0 \]
Factoring out the common factors in each group:
\[ 2x(3x - 2) + 1(3x - 2) = 0 \]

### Step 5: Factor out the common binomial
Now we can factor out the common binomial \( (3x - 2) \):
\[ (3x - 2)(2x + 1) = 0 \]

### Step 6: Set each factor to zero
Now we set each factor equal to zero:
1. \( 3x - 2 = 0 \)
2. \( 2x + 1 = 0 \)

### Step 7: Solve for \( x \)
For the first equation:
\[ 3x - 2 = 0 \implies 3x = 2 \implies x = \frac{2}{3} \]

For the second equation:
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]

### Step 8: Write the solutions
The solutions to the equation \( 6x^2 - x - 2 = 0 \) are:
\[ x = -\frac{1}{2}, \quad x = \frac{2}{3} \]

### Summary of the Solutions
The roots of the quadratic equation are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = \frac{2}{3} \]

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