(y²+2y+4)²-7(y²+2y+4)+12=0

To solve the equation \((y^2 + 2y + 4)^2 - 7(y^2 + 2y + 4) + 12 = 0\), let's first introduce a substitution to simplify the expression. Let:

\[ x = y^2 + 2y + 4 \]

Now, we can rewrite the equation in terms of \(x\):

\[ x^2 - 7x + 12 = 0 \]

Next, we will factor the quadratic equation. We want to find two numbers that multiply to \(12\) and add to \(-7\). These numbers are \(-3\) and \(-4\). Therefore, we can factor the equation as:

\[ (x - 3)(x - 4) = 0 \]

Setting each factor to zero gives:

\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

Now, we substitute back \(x = y^2 + 2y + 4\).

Case 1: \(y^2 + 2y + 4 = 3\)

\[ y^2 + 2y + 4 - 3 = 0 \quad \Rightarrow \quad y^2 + 2y + 1 = 0 \] \[ (y + 1)^2 = 0 \quad \Rightarrow \quad y + 1 = 0 \quad \Rightarrow \quad y = -1 \]

Case 2: \(y^2 + 2y + 4 = 4\)

\[ y^2 + 2y + 4 - 4 = 0 \quad \Rightarrow \quad y^2 + 2y = 0 \] \[ y(y + 2) = 0 \] \[ y = 0 \quad \text{or} \quad y + 2 = 0 \quad \Rightarrow \quad y = -2 \]

Final Solutions

So, the solutions to the original equation are:

\[ y = -1, \quad y = 0, \quad y = -2 \]

These are the final results!