We are given the equation:

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

Let’s solve it step by step.

### Step 1: Rearrange the equation
We want to set the equation to 0, so move all terms to one side:

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

This is a quadratic equation of the form \( ay^2 + by + c = 0 \), where \( a = 6 \), \( b = -3 \), and \( c = 5 \).

### Step 2: Use the quadratic formula
The quadratic formula is given by:

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

Substitute \( a = 6 \), \( b = -3 \), and \( c = 5 \):

\[
y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(6)(5)}}{2(6)}
\]

\[
y = \frac{3 \pm \sqrt{9 - 120}}{12}
\]

\[
y = \frac{3 \pm \sqrt{-111}}{12}
\]

### Step 3: Handle the discriminant
Since the discriminant \( \sqrt{-111} \) is negative, the solutions will be complex numbers. We can express the solution as:

\[
y = \frac{3 \pm i\sqrt{111}}{12}
\]

### Final Answer:
The solutions are:

\[
y = \frac{3 + i\sqrt{111}}{12} \quad \text{and} \quad y = \frac{3 - i\sqrt{111}}{12}
\]

These are the complex solutions to the quadratic equation.

Would you like a detailed breakdown of any step, or do you have any specific questions?

Here are 5 related questions to explore further:
1. How do you calculate the discriminant of a quadratic equation?
2. What are complex numbers, and how are they represented?
3. Can you graph a quadratic equation with complex solutions?
4. How do imaginary numbers affect the roots of a quadratic equation?
5. How does the quadratic formula work in cases of real solutions?

**Tip:** When solving quadratic equations, always check the discriminant \( b^2 - 4ac \) to determine if the solutions will be real or complex.

Discover how to solve the quadratic equation -3y + 5 = -6y^2 step-by-step, including a discussion on the discriminant and complex solutions. Perfect for students in Grades 10-12, this problem explores quadratic equations, complex numbers, and algebraic formulas.

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