To solve the quadratic equation \(6v^2 + 19v - 104 = 8v - 9\), follow these steps:

### Step 1: Simplify the equation
First, move all terms to one side of the equation by subtracting \(8v\) and adding \(9\) to both sides:

\[
6v^2 + 19v - 104 - 8v + 9 = 0
\]

This simplifies to:

\[
6v^2 + 11v - 95 = 0
\]

### Step 2: Solve the quadratic equation
Now, we can solve the quadratic equation using the quadratic formula:

\[
v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \(6v^2 + 11v - 95 = 0\), the coefficients are:
- \(a = 6\)
- \(b = 11\)
- \(c = -95\)

### Step 3: Calculate the discriminant
The discriminant (\(\Delta\)) is given by:

\[
\Delta = b^2 - 4ac = 11^2 - 4(6)(-95)
\]

\[
\Delta = 121 + 2280 = 2401
\]

### Step 4: Find the roots
Since the discriminant is positive, there are two real roots:

\[
v = \frac{-11 \pm \sqrt{2401}}{2(6)} = \frac{-11 \pm 49}{12}
\]

Calculating the two roots:

1. \(v_1 = \frac{-11 + 49}{12} = \frac{38}{12} = \frac{19}{6}\)
2. \(v_2 = \frac{-11 - 49}{12} = \frac{-60}{12} = -5\)

### Final Answer:
The solutions to the equation are:

\[
v = \frac{19}{6} \quad \text{or} \quad v = -5
\]

Would you like more details or have any questions?

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Here are some related questions to consider:
1. How do you determine the discriminant in a quadratic equation?
2. What does the discriminant tell you about the roots of a quadratic equation?
3. How can you verify the solutions to a quadratic equation?
4. What are the possible types of solutions for a quadratic equation?
5. How can the quadratic formula be derived?
6. What happens if the discriminant is zero?
7. What if the discriminant is negative?
8. Can you solve a quadratic equation by factoring instead of using the quadratic formula?

**Tip:** Always simplify the equation first by moving all terms to one side before applying any solving techniques.

Learn how to solve the quadratic equation 6v^2 + 19v - 104 = 8v - 9 with a detailed step-by-step solution. This problem covers key algebraic concepts, including the quadratic formula and discriminant. Suitable for students in Grades 9-12.

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