Reduce the equation `y(2y + 15) = 3(y^(2) + y + 8)` to the standard form.

Answer

To reduce the equation \( y(2y + 15) = 3(y^2 + y + 8) \) to standard form, we will follow these steps:

### Step 1: Expand both sides of the equation
Start by distributing the terms on both sides of the equation.

\[ y(2y + 15) = 3(y^2 + y + 8) \]

Expanding the left side:
\[ y \cdot 2y + y \cdot 15 = 2y^2 + 15y \]

Expanding the right side:
\[ 3 \cdot y^2 + 3 \cdot y + 3 \cdot 8 = 3y^2 + 3y + 24 \]

Now we rewrite the equation:
\[ 2y^2 + 15y = 3y^2 + 3y + 24 \]

### Step 2: Move all terms to one side
Next, we will move all terms to one side of the equation to set it to zero.

Subtract \(3y^2\), \(3y\), and \(24\) from both sides:
\[ 2y^2 + 15y - 3y^2 - 3y - 24 = 0 \]

### Step 3: Combine like terms
Now, combine the like terms:
\[ (2y^2 - 3y^2) + (15y - 3y) - 24 = 0 \]
This simplifies to:
\[ -y^2 + 12y - 24 = 0 \]

### Step 4: Multiply through by -1
To write the equation in standard form \( ax^2 + bx + c = 0 \), we multiply the entire equation by -1:
\[ y^2 - 12y + 24 = 0 \]

### Final Result
The equation in standard form is:
\[ y^2 - 12y + 24 = 0 \]

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