Find the value of X and Y (X और Y का मान ज्ञात कीजिए)
I. `2x^(2)-5x-3=0`
II. `4y^(2)+16y+15=0`

Options

A. `xgty`
B. `xlty`
C. `xley`
D. `x=y`

Answer

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve for X in the equation \(2x^2 - 5x - 3 = 0\)

1. **Identify the coefficients**:
- \(a = 2\), \(b = -5\), \(c = -3\).

2. **Use the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot (-3) = 25 + 24 = 49 \]

4. **Substitute into the quadratic formula**:
\[ x = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4} \]

5. **Calculate the two possible values for x**:
- \(x_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3\)
- \(x_2 = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}\)

### Step 2: Solve for Y in the equation \(4y^2 + 16y + 15 = 0\)

1. **Identify the coefficients**:
- \(a = 4\), \(b = 16\), \(c = 15\).

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (16)^2 - 4 \cdot 4 \cdot 15 = 256 - 240 = 16 \]

4. **Substitute into the quadratic formula**:
\[ y = \frac{-16 \pm \sqrt{16}}{8} = \frac{-16 \pm 4}{8} \]

5. **Calculate the two possible values for y**:
- \(y_1 = \frac{-16 + 4}{8} = \frac{-12}{8} = -\frac{3}{2}\)
- \(y_2 = \frac{-16 - 4}{8} = \frac{-20}{8} = -\frac{5}{2}\)

### Final Results:
- The values of \(X\) are \(3\) and \(-\frac{1}{2}\).
- The values of \(Y\) are \(-\frac{3}{2}\) and \(-\frac{5}{2}\).

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