Let's solve the system of equations:

1. \( x^2 + 2y - 12 = 0 \)  (Equation 1)
2. \( x^2 + y^2 = 36 \)    (Equation 2)

### Step 1: Express \( x^2 \) from Equation 1

From Equation 1:
\[
x^2 = 12 - 2y \quad \text{(Equation 3)}
\]

### Step 2: Substitute \( x^2 \) in Equation 2

Substituting Equation 3 into Equation 2:
\[
(12 - 2y) + y^2 = 36
\]

Simplify:
\[
y^2 - 2y + 12 = 36
\]

\[
y^2 - 2y - 24 = 0
\]

### Step 3: Solve the quadratic equation for \( y \)

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-24)}}{2(1)}
\]

\[
y = \frac{2 \pm \sqrt{4 + 96}}{2}
\]

\[
y = \frac{2 \pm \sqrt{100}}{2}
\]

\[
y = \frac{2 \pm 10}{2}
\]

\[
y = 6 \quad \text{or} \quad y = -4
\]

### Step 4: Find corresponding \( x \) values

For \( y = 6 \):
\[
x^2 = 12 - 2(6) = 0 \implies x = 0
\]

For \( y = -4 \):
\[
x^2 = 12 - 2(-4) = 20 \implies x = \pm \sqrt{20} = \pm 2\sqrt{5}
\]

### Solutions

The solutions are:
- \( (x, y) = (0, 6) \)
- \( (x, y) = (2\sqrt{5}, -4) \)
- \( (x, y) = (-2\sqrt{5}, -4) \)

Would you like more details or have any questions?

Here are some related questions you might explore:

1. How do you solve systems of nonlinear equations in general?
2. What are the implications of having multiple solutions in such systems?
3. How can you verify the solutions to a system of equations?
4. What methods can be used if one equation is not easily solvable for a variable?
5. How do changes in coefficients affect the number and type of solutions?

**Tip:** Always check your solutions by substituting them back into the original equations to ensure they satisfy all given conditions.

Learn how to solve the system of nonlinear equations x^2 + 2y - 12 = 0 and x^2 + y^2 = 36 with a step-by-step solution. Explore key algebraic concepts, such as substitution, quadratic equations, and the quadratic formula. Suitable for high school students in Grades 9-12.

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