Let's solve the equation step by step.

The given equation is:

\[
20.05y + 28y + 28y \cdot 1.083y = 3.89 + 3.89 \cdot 1.083y
\]

### Step 1: Simplify the equation
First, simplify the terms on both sides.

#### Left Side:
Combine like terms:
\[
20.05y + 28y + 28y \cdot 1.083y
\]
\[
(20.05 + 28)y + 28y \cdot 1.083y
\]
\[
48.05y + 28y \cdot 1.083y
\]

#### Right Side:
Similarly, simplify the right side:
\[
3.89 + 3.89 \cdot 1.083y
\]
This simplifies to:
\[
3.89(1 + 1.083y)
\]

### Step 2: Write the equation
So, the equation becomes:
\[
48.05y + 28y \cdot 1.083y = 3.89(1 + 1.083y)
\]

### Step 3: Distribute and simplify further

The next step is to expand and move all terms involving \( y \) to one side:

Expand the right side:
\[
48.05y + 28y \cdot 1.083y = 3.89 + 4.21347y
\]

You now have:
\[
48.05y + 30.324y^2 = 3.89 + 4.21347y
\]

### Step 4: Rearrange the equation
Move all terms to one side to set the equation to zero:
\[
30.324y^2 + (48.05y - 4.21347y) - 3.89 = 0
\]

Simplify further:
\[
30.324y^2 + 43.83653y - 3.89 = 0
\]

### Step 5: Solve the quadratic equation
This is now a quadratic equation of the form:
\[
ay^2 + by + c = 0
\]
where \( a = 30.324 \), \( b = 43.83653 \), and \( c = -3.89 \).

Using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substituting the values:
\[
y = \frac{-43.83653 \pm \sqrt{(43.83653)^2 - 4(30.324)(-3.89)}}{2(30.324)}
\]

### Step 6: Calculate the solution

Let's calculate the value of \( y \).The solutions to the quadratic equation are:

\[
y_1 = 0.0839 \quad \text{(approximately)}
\]
\[
y_2 = -1.5295 \quad \text{(approximately)}
\]

So, the equation has two possible solutions for \( y \):

1. \( y \approx 0.0839 \)
2. \( y \approx -1.5295 \)

Would you like more details on any step, or do you have any questions?

### Related Questions:
1. How do you determine which solution to use in a real-world context?
2. What is the significance of the discriminant in a quadratic equation?
3. How can you check which solution is valid for a particular problem?
4. What happens if the discriminant is negative?
5. How do you simplify quadratic expressions before applying the quadratic formula?
6. What are the implications of having two solutions in a quadratic equation?
7. How can you graphically interpret the solutions of a quadratic equation?
8. What are some applications of quadratic equations in physics?

**Tip:** Always consider the context of the problem when choosing which solution of a quadratic equation to use, especially if the variable represents a physical quantity.

Learn how to solve the equation 20.05y + 28y + 28y * 1.083y = 3.89 + 3.89 * 1.083y with a detailed step-by-step solution. This problem covers key algebraic concepts, including solving equations with multiplication of terms. Suitable for advanced high school students and beyond.

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