Let's first solve the quadratic equation $x^2 + 7x + 12 = 0$ and then apply the result to solve the second equation.

Step 1: Solve $x^2 + 7x + 12 = 0$

We solve this quadratic equation using factoring: $x^2 + 7x + 12 = 0$ We need two numbers that multiply to 12 and add up to 7. These numbers are 3 and 4.

Thus, we factor the quadratic as: $(x + 3)(x + 4) = 0$

Setting each factor to zero gives: $x + 3 = 0 \quad \text{or} \quad x + 4 = 0$ $x = -3 \quad \text{or} \quad x = -4$

So, the solutions for $x^2 + 7x + 12 = 0$ are: $x = -3 \quad \text{or} \quad x = -4$

Step 2: Solve $(y^2 + 4y)^2 + 7(y^2 + 4y) + 12 = 0$

Now, let $z = y^2 + 4y$. This transforms the equation into: $z^2 + 7z + 12 = 0$

This is the same quadratic equation as before, so using the factorization, we have: $(z + 3)(z + 4) = 0$

Thus, $z = -3$ or $z = -4$.

Step 3: Solve for $y$

Now, substitute $z = y^2 + 4y$:

1. For $y^2 + 4y = -3$: $y^2 + 4y + 3 = 0$ Factor this as: $(y + 1)(y + 3) = 0$ So, $y = -1$ or $y = -3$.

2. For $y^2 + 4y = -4$: $y^2 + 4y + 4 = 0$ Factor this as: $(y + 2)^2 = 0$ So, $y = -2$.

Final Solutions

The solutions to the second equation are: $y = -1, \quad y = -3, \quad y = -2$

Would you like further details or explanations? Here are some related questions:

1. How do you solve a quadratic equation using the quadratic formula?
2. What are the different methods of factoring quadratics?
3. Can we always solve quadratic equations by factoring?
4. How does substitution help in solving more complex equations?
5. What are the applications of quadratic equations in real life?

Tip: When solving quadratic equations, always check if factoring is possible before using more complex methods like the quadratic formula.