We are given the equation $x^2 + xy + y^2 = 6$. We are asked to solve this equation, but it is not stated what variable to solve for. I will assume that the goal is to find all real solutions $(x, y)$ to the equation $x^2 + xy + y^2 = 6$.

1. Problem Description

We are given the equation $x^2 + xy + y^2 = 6$. We are asked to solve this equation, but it is not stated what variable to solve for.

I will assume that the goal is to find all real solutions $(x, y)$ to the equation $x^2 + xy + y^2 = 6$.

2. Solution Steps

We can analyze this equation in different ways.

First, let's try to consider $y$ as a function of $x$. We can rewrite the given equation as a quadratic equation in terms of $y$:

$y^2 + xy + (x^2 - 6) = 0$

Using the quadratic formula to solve for $y$, we have:

$y = \frac{-x \pm \sqrt{x^2 - 4(x^2 - 6)}}{2}$

$y = \frac{-x \pm \sqrt{x^2 - 4x^2 + 24}}{2}$

$y = \frac{-x \pm \sqrt{24 - 3x^2}}{2}$

For $y$ to be real, the expression under the square root must be non-negative:

$24 - 3x^2 \geq 0$

$3x^2 \leq 24$

$x^2 \leq 8$

$-\sqrt{8} \leq x \leq \sqrt{8}$

$-2\sqrt{2} \leq x \leq 2\sqrt{2}$

Now we can see that for each value of $x$ within the interval $[-2\sqrt{2}, 2\sqrt{2}]$, there are two possible values for $y$:

$y = \frac{-x + \sqrt{24 - 3x^2}}{2}$

$y = \frac{-x - \sqrt{24 - 3x^2}}{2}$

For example, when $x=0$, we have $y = \frac{\pm\sqrt{24}}{2} = \frac{\pm 2\sqrt{6}}{2} = \pm \sqrt{6}$.

When $x = 2\sqrt{2}$, $y = \frac{-2\sqrt{2} \pm \sqrt{24 - 3(8)}}{2} = \frac{-2\sqrt{2} \pm 0}{2} = -\sqrt{2}$.

When $x = -2\sqrt{2}$, $y = \frac{2\sqrt{2} \pm \sqrt{24 - 3(8)}}{2} = \frac{2\sqrt{2} \pm 0}{2} = \sqrt{2}$.

The solution set consists of all pairs $(x, y)$ such that $-2\sqrt{2} \leq x \leq 2\sqrt{2}$ and $y = \frac{-x \pm \sqrt{24 - 3x^2}}{2}$.

The solution set can also be described as the set of all pairs $(x, \frac{-x \pm \sqrt{24-3x^2}}{2})$ for $-2\sqrt{2} \leq x \leq 2\sqrt{2}$.

3. Final Answer

The solution to the equation $x^2 + xy + y^2 = 6$ is given by the set of all pairs $(x, y)$ such that $-2\sqrt{2} \leq x \leq 2\sqrt{2}$ and $y = \frac{-x \pm \sqrt{24 - 3x^2}}{2}$.