To solve for \( y \) in the equation:

\[
a = 3w^2 + 4wy + y^2
\]

we recognize this as a quadratic equation in \( y \). Rearranging the terms:

\[
y^2 + 4wy + (3w^2 - a) = 0
\]

This is now a quadratic equation in the standard form \( Ay^2 + By + C = 0 \), where:
- \( A = 1 \)
- \( B = 4w \)
- \( C = 3w^2 - a \)

We can apply the quadratic formula to solve for \( y \):

\[
y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]

Substitute the values of \( A \), \( B \), and \( C \):

\[
y = \frac{-4w \pm \sqrt{(4w)^2 - 4(1)(3w^2 - a)}}{2(1)}
\]

Simplify the expression:

\[
y = \frac{-4w \pm \sqrt{16w^2 - 4(3w^2 - a)}}{2}
\]
\[
y = \frac{-4w \pm \sqrt{16w^2 - 12w^2 + 4a}}{2}
\]
\[
y = \frac{-4w \pm \sqrt{4w^2 + 4a}}{2}
\]
\[
y = \frac{-4w \pm 2\sqrt{w^2 + a}}{2}
\]

Finally, simplify the result:

\[
y = -2w \pm \sqrt{w^2 + a}
\]

Thus, the two possible solutions for \( y \) are:

\[
y = -2w + \sqrt{w^2 + a} \quad \text{or} \quad y = -2w - \sqrt{w^2 + a}
\]

Would you like further clarification on any part of this solution?

Here are 5 related questions you might explore:
1. How do you identify when to use the quadratic formula?
2. What are the implications of the discriminant in this equation?
3. Can the quadratic formula give complex solutions in some cases?
4. How do you simplify square root expressions in quadratic equations?
5. What is the significance of completing the square in solving quadratics?

**Tip:** When applying the quadratic formula, ensure that the discriminant (inside the square root) is evaluated carefully to check for real vs complex solutions.

This guide solves the quadratic equation a = 3w² + 4wy + y² to find y in terms of a and w. Learn the step-by-step process using the quadratic formula and simplifying the result. Suitable for high school students in Grades 9-11, this problem highlights key concepts in algebra and quadratic equations.

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