Let's solve the quadratic equation \( 5.2y^2 - 3 = 5y \) by completing the square. Here are the steps:

### Step 1: Rearrange the equation
Move all terms to one side of the equation:
\[
5.2y^2 - 5y - 3 = 0
\]

### Step 2: Divide the equation by 5.2
To simplify, divide through by 5.2:
\[
y^2 - \frac{5}{5.2}y - \frac{3}{5.2} = 0
\]
Simplifying the fractions:
\[
y^2 - 0.9615y - 0.5769 = 0
\]

### Step 3: Move the constant term to the other side
Move the constant term to the other side:
\[
y^2 - 0.9615y = 0.5769
\]

### Step 4: Complete the square
To complete the square, take half of the coefficient of \( y \), square it, and add it to both sides:
\[
\left(\frac{-0.9615}{2}\right)^2 = 0.2309^2 = 0.0533
\]
Add this to both sides:
\[
y^2 - 0.9615y + 0.0533 = 0.5769 + 0.0533
\]
\[
(y - 0.48075)^2 = 0.6302
\]

### Step 5: Solve for \( y \)
Take the square root of both sides:
\[
y - 0.48075 = \pm \sqrt{0.6302}
\]
\[
y - 0.48075 = \pm 0.7938
\]

### Step 6: Find the two solutions
Now solve for \( y \):
\[
y = 0.48075 + 0.7938 \quad \text{or} \quad y = 0.48075 - 0.7938
\]
\[
y = 1.27455 \quad \text{or} \quad y = -0.31305
\]

### Final Solution:
The solutions are approximately:
\[
y \approx 1.27 \quad \text{or} \quad y \approx -0.31
\]

Would you like further clarifications on any step?

### Relative Questions:
1. How do you simplify quadratic equations with decimal coefficients?
2. Can completing the square always solve any quadratic equation?
3. How do you know when to complete the square versus using other methods like factoring?
4. What happens when the number inside the square root is negative?
5. How does completing the square help in deriving the quadratic formula?

**Tip:** When dealing with decimal coefficients, you can multiply through by a factor to eliminate decimals, making calculations easier.

Learn how to solve the quadratic equation 5.2y² - 5y - 3 step-by-step using completing the square method. This problem involves decimal coefficients and covers key algebraic concepts, including the quadratic equation formula and completing the square method. Suitable for students in Grades 10-12.

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