Solve for x x^2+xy+y^2=0

$x^{2} + x y + y^{2} = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = y$ , and $c = y^{2}$ into the quadratic formula and solve for $x$ .

$\frac{- y \pm \sqrt{y^{2} - 4 \cdot (1 \cdot y^{2})}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Factor $y^{2}$ out of $y^{2} - 4 \cdot 1 \cdot y^{2}$ .

Multiply by $1$ .

$x = \frac{- y \pm \sqrt{y^{2} \cdot 1 - 4 \cdot 1 \cdot y^{2}}}{2 \cdot 1}$

Factor $y^{2}$ out of $- 4 \cdot 1 \cdot y^{2}$ .

$x = \frac{- y \pm \sqrt{y^{2} \cdot 1 + y^{2} \cdot (- 4 \cdot 1)}}{2 \cdot 1}$

Factor $y^{2}$ out of $y^{2} \cdot 1 + y^{2} \cdot (- 4 \cdot 1)$ .

$x = \frac{- y \pm \sqrt{y^{2} (1 - 4 \cdot 1)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- y \pm \sqrt{y^{2} (1 - 4)}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{- y \pm \sqrt{y^{2} \cdot - 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- y \pm y \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- y \pm y \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- y \pm y (\sqrt{- 1} \cdot \sqrt{3})}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- y \pm y i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

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

The final answer is the combination of both solutions.

$x = - \frac{y (1 - i \sqrt{3})}{2}$

$x = - \frac{y (1 + i \sqrt{3})}{2}$