Solve the equation for all real solutions in simplest form.

3, z, squared, minus, z, plus, 5, equals, 6

3*z*2−*z*+5=

6




Let's solve the equation for \( z \):

\[
3z^2 - z + 5 = 6
\]

### Step 1: Simplify the Equation
First, move all terms to one side of the equation to set it equal to zero:

\[
3z^2 - z + 5 - 6 = 0
\]

\[
3z^2 - z - 1 = 0
\]

### Step 2: Use the Quadratic Formula
This is a quadratic equation of the form \( az^2 + bz + c = 0 \), where:
- \( a = 3 \)
- \( b = -1 \)
- \( c = -1 \)

The quadratic formula is:

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

Substitute the values for \( a \), \( b \), and \( c \):

\[
z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}
\]

\[
z = \frac{1 \pm \sqrt{1 + 12}}{6}
\]

\[
z = \frac{1 \pm \sqrt{13}}{6}
\]

### Step 3: Simplify the Solution
The solutions are:

\[
z = \frac{1 + \sqrt{13}}{6} \quad \text{and} \quad z = \frac{1 - \sqrt{13}}{6}
\]

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

\[
z = \frac{1 + \sqrt{13}}{6} \quad \text{and} \quad z = \frac{1 - \sqrt{13}}{6}
\]

Would you like further details or have any questions?

---

1. How do we derive the quadratic formula?
2. What are real and complex solutions in quadratic equations?
3. How can we check if our solutions satisfy the original equation?
4. What are the steps to simplify square roots in solutions?
5. How would this process differ if the discriminant were negative?

**Tip:** For quadratic equations, if the discriminant (\(b^2 - 4ac\)) is positive, there are two real solutions; if it's zero, there's one; if negative, the solutions are complex.

Solve the equation for all real solutions in simplest form: 3z^2 - z + 5 = 6

Step-by-step guide to solving the quadratic equation 3z^2 - z + 5 = 6 using the quadratic formula. This solution covers essential algebraic concepts and provides detailed explanations for each step. Ideal for high school students in Grades 9-12.

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