If `A -={x//6 x ^(2)+ x-15=0},`
`B-={x//2x ^(2) - 5x +3 =0}and `
`C-= {x//2x ^(2) -x-3 =0},` then `a nn B nnC=`

Options

A. `{-(5)/(3), (3)/(2)}`
B. `{1,(3)/(2)}`
C. `{-1,(3)/(2)}`
D. `{(3)/(2)}`

Answer

To solve the problem, we need to find the elements of sets A, B, and C based on the given quadratic equations, and then determine the intersection of these sets.

### Step 1: Find the elements of Set A
Set A is defined by the equation:
\[ 6x^2 + x - 15 = 0 \]

We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
where \( D = b^2 - 4ac \).

Here, \( a = 6 \), \( b = 1 \), and \( c = -15 \).

1. Calculate \( D \):
\[ D = 1^2 - 4 \cdot 6 \cdot (-15) = 1 + 360 = 361 \]

2. Now, apply the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{361}}{2 \cdot 6} = \frac{-1 \pm 19}{12} \]

3. Calculate the two possible values of \( x \):
- First value:
\[ x_1 = \frac{-1 + 19}{12} = \frac{18}{12} = \frac{3}{2} \]
- Second value:
\[ x_2 = \frac{-1 - 19}{12} = \frac{-20}{12} = \frac{-5}{3} \]

Thus, the elements of Set A are:
\[ A = \left\{ \frac{3}{2}, -\frac{5}{3} \right\} \]

### Step 2: Find the elements of Set B
Set B is defined by the equation:
\[ 2x^2 - 5x + 3 = 0 \]

Using the quadratic formula:
Here, \( a = 2 \), \( b = -5 \), and \( c = 3 \).

1. Calculate \( D \):
\[ D = (-5)^2 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1 \]

2. Apply the quadratic formula:
\[ x = \frac{5 \pm \sqrt{1}}{2 \cdot 2} = \frac{5 \pm 1}{4} \]

3. Calculate the two possible values of \( x \):
- First value:
\[ x_1 = \frac{6}{4} = \frac{3}{2} \]
- Second value:
\[ x_2 = \frac{4}{4} = 1 \]

Thus, the elements of Set B are:
\[ B = \left\{ \frac{3}{2}, 1 \right\} \]

### Step 3: Find the elements of Set C
Set C is defined by the equation:
\[ 2x^2 - x - 3 = 0 \]

Using the quadratic formula:
Here, \( a = 2 \), \( b = -1 \), and \( c = -3 \).

1. Calculate \( D \):
\[ D = (-1)^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25 \]

2. Apply the quadratic formula:
\[ x = \frac{1 \pm \sqrt{25}}{2 \cdot 2} = \frac{1 \pm 5}{4} \]

3. Calculate the two possible values of \( x \):
- First value:
\[ x_1 = \frac{6}{4} = \frac{3}{2} \]
- Second value:
\[ x_2 = \frac{-4}{4} = -1 \]

Thus, the elements of Set C are:
\[ C = \left\{ \frac{3}{2}, -1 \right\} \]

### Step 4: Find the intersection of sets A, B, and C
Now we need to find \( A \cap B \cap C \):
- Set A: \( \left\{ \frac{3}{2}, -\frac{5}{3} \right\} \)
- Set B: \( \left\{ \frac{3}{2}, 1 \right\} \)
- Set C: \( \left\{ \frac{3}{2}, -1 \right\} \)

The common element in all three sets is:
\[ \frac{3}{2} \]

### Final Answer
Thus, the intersection \( A \cap B \cap C \) is:
\[ A \cap B \cap C = \left\{ \frac{3}{2} \right\} \]

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