Solve the equation for all real solutions in simplest form.

2, v, squared, plus, 15, v, plus, 19, equals, 0

2*v*2+15*v*+19=

0

To solve the quadratic equation \( 2v^2 + 15v + 19 = 0 \), we can use the quadratic formula:

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

where \( a = 2 \), \( b = 15 \), and \( c = 19 \).

### Step 1: Plugging values into the quadratic formula
\[
v = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 2 \cdot 19}}{2 \cdot 2}
\]

### Step 2: Simplifying under the square root
\[
v = \frac{-15 \pm \sqrt{225 - 152}}{4}
\]
\[
v = \frac{-15 \pm \sqrt{73}}{4}
\]

### Step 3: Writing the final answer
The solutions are:
\[
v = \frac{-15 + \sqrt{73}}{4} \quad \text{and} \quad v = \frac{-15 - \sqrt{73}}{4}
\]

These are the two real solutions in their simplest form.

Would you like further details on any step?

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1. How can we determine if a quadratic has real solutions?
2. What does the discriminant tell us in quadratic equations?
3. How do complex solutions arise in quadratic equations?
4. What are other methods to solve quadratics besides the quadratic formula?
5. How does completing the square solve quadratics?

**Tip:** The discriminant (\( b^2 - 4ac \)) indicates the number and type of solutions. If positive, there are two real solutions; if zero, one real solution; if negative, two complex solutions.

Solve the equation for all real solutions in simplest form.

2v^2 + 15v + 19 = 0

Discover how to solve the quadratic equation 2v^2 + 15v + 19 = 0 using the quadratic formula. Follow a step-by-step solution that simplifies the problem to find all real solutions, covering key concepts in algebra and quadratic equations, suitable for students in Grades 9-11.

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