Solve the equation for all real solutions in simplest form.
3, v, squared, plus, 3, v, minus, 4, equals, 0
3v 
2
 +3v−4=
0

The given equation is:

\[
3v^2 + 3v - 4 = 0
\]

We can solve this quadratic equation using the quadratic formula. The quadratic formula is given by:

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

In our equation, we have:
- \( a = 3 \)
- \( b = 3 \)
- \( c = -4 \)

Now, let's plug these values into the quadratic formula:

\[
v = \frac{-3 \pm \sqrt{3^2 - 4(3)(-4)}}{2(3)}
\]

First, calculate the discriminant:

\[
\Delta = b^2 - 4ac = 3^2 - 4(3)(-4) = 9 + 48 = 57
\]

Now substitute the discriminant back into the quadratic formula:

\[
v = \frac{-3 \pm \sqrt{57}}{6}
\]

Thus, the solutions are:

\[
v = \frac{-3 + \sqrt{57}}{6} \quad \text{or} \quad v = \frac{-3 - \sqrt{57}}{6}
\]

These are the two real solutions in simplest form.

Would you like further clarification on any of the steps? Here are five related questions:

1. How do you derive the quadratic formula?
2. What does the discriminant tell us about the nature of the solutions?
3. Can a quadratic equation have no real solutions?
4. How do you complete the square to solve quadratic equations?
5. What are some examples of quadratic equations with integer solutions?

**Tip:** The discriminant \( \Delta = b^2 - 4ac \) can tell you whether a quadratic equation has real or complex solutions. If \( \Delta > 0 \), there are two real solutions.

Solve the equation for all real solutions in simplest form: 3v^2 + 3v - 4 = 0

This step-by-step guide explains how to solve the quadratic equation 3v^2 + 3v - 4 = 0 using the quadratic formula. Includes an explanation of the discriminant and its role in determining the number of real solutions. Ideal for students in Grades 9-11.

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