solve the equation for y. -x^2+y^2+8x-16=0. Enter the expression(s) for 
y​. Separate multiple answers with commas, if necessary.

To solve the equation for \( y \), let's proceed step by step.

The given equation is:

\[
-x^2 + y^2 + 8x - 16 = 0
\]

1. Rearrange the equation to isolate \( y^2 \) on one side:

\[
y^2 = x^2 - 8x + 16
\]

2. Now, take the square root of both sides to solve for \( y \):

\[
y = \pm \sqrt{x^2 - 8x + 16}
\]

3. Notice that \( x^2 - 8x + 16 \) is a perfect square trinomial, which can be factored as:

\[
y = \pm \sqrt{(x - 4)^2}
\]

4. Simplifying the square root:

\[
y = \pm (x - 4)
\]

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

\[
y = x - 4, \quad y = -(x - 4)
\]

Or, equivalently:

\[
y = x - 4, \quad y = -x + 4
\]

These are the two expressions for \( y \).

---

Would you like further details or have any questions?

Here are some related questions for practice:

1. Solve for \( y \) in the equation \( x^2 + y^2 - 2x + 4 = 0 \).
2. Solve \( y^2 - 6x + 9 = 0 \) for \( y \).
3. Factor the quadratic expression \( x^2 - 10x + 25 \).
4. Find the vertex of the parabola described by \( y = x^2 - 8x + 16 \).
5. Solve the system of equations \( y = x - 3 \) and \( y^2 + x^2 = 25 \).

**Tip:** Recognizing perfect square trinomials can make factoring and solving quadratic equations much faster.

solve the equation for y. -x^2 + y^2 + 8x - 16 = 0. Enter the expression(s) for y​. Separate multiple answers with commas, if necessary.

Learn how to solve for y in the equation -x^2 + y^2 + 8x - 16 = 0. This problem involves recognizing a perfect square trinomial and solving quadratic equations, suitable for Grades 9-10. Follow step-by-step instructions and get multiple solutions for y.

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