`3x ^(2)=m (5x +v)`
What are the values of x that satisgy the equation above, where m and v are constants ?

Options

A. `x =- (5m )/(6) pm (sqrt(25m ^(2)+12mv))/(6)`
B. `x = (5m )/(6) pm (sqrt(25m ^(2)+12mv))/(6)`
C. `x =- (5m )/(3) pm (sqrt(12m ^(2)+12mv))/(3)`
D. `x = (5m )/(3) pm (sqrt(25m ^(2)+12mv))/(3)`

Explanation

To solve the equation \( 3x^2 = m(5x + v) \), we will rearrange it into the standard form of a quadratic equation and then apply the quadratic formula to find the values of \( x \).

### Step-by-Step Solution:

1. **Rearranging the Equation:**
Start with the given equation:
\[ 3x^2 = m(5x + v) \]
Distribute \( m \) on the right side:
\[ 3x^2 = 5mx + mv \]
Now, rearranging this to bring all terms to one side gives:
\[ 3x^2 - 5mx - mv = 0 \]

2. **Identifying Coefficients:**
The standard form of a quadratic equation is:
\[ ax^2 + bx + c = 0 \]
From our equation \( 3x^2 - 5mx - mv = 0 \), we can identify:
- \( a = 3 \)
- \( b = -5m \)
- \( c = -mv \)

3. **Applying the Quadratic Formula:**
The quadratic formula for finding the roots of the equation \( ax^2 + bx + c = 0 \) is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in our coefficients:
\[ x = \frac{-(-5m) \pm \sqrt{(-5m)^2 - 4 \cdot 3 \cdot (-mv)}}{2 \cdot 3} \]
Simplifying this gives:
\[ x = \frac{5m \pm \sqrt{25m^2 + 12mv}}{6} \]

4. **Final Expression for x:**
Thus, the values of \( x \) that satisfy the equation are:
\[ x = \frac{5m}{6} \pm \frac{\sqrt{25m^2 + 12mv}}{6} \]