Solve x^2+x+y^2=(x+2y)^2

$$x ^ { 2 } + x + y ^ { 2 } = ( x + 2 y ) ^ { 2 }$$

Solve for x

$$x=\frac{3y^{2}}{1-4y}$$

$$y\neq \frac{1}{4}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2y\right)^{2}.

$$x^{2}+x+y^{2}=x^{2}+4xy+4y^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+x+y^{2}-x^{2}=4xy+4y^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$x+y^{2}=4xy+4y^{2}$$

Subtract 4xy from both sides.

$$x+y^{2}-4xy=4y^{2}$$

Subtract y^{2} from both sides.

$$x-4xy=4y^{2}-y^{2}$$

Combine 4y^{2} and -y^{2} to get 3y^{2}.

$$x-4xy=3y^{2}$$

Combine all terms containing x.

$$\left(1-4y\right)x=3y^{2}$$

Divide both sides by -4y+1.

$$\frac{\left(1-4y\right)x}{1-4y}=\frac{3y^{2}}{1-4y}$$

Dividing by -4y+1 undoes the multiplication by -4y+1.

$$x=\frac{3y^{2}}{1-4y}$$

Solve for y (complex solution)

$$y=\frac{-\sqrt{x\left(4x+3\right)}-2x}{3}$$

$$y=\frac{\sqrt{x\left(4x+3\right)}-2x}{3}$$

Solve for y

$$y=\frac{-\sqrt{x\left(4x+3\right)}-2x}{3}$$

$$y=\frac{\sqrt{x\left(4x+3\right)}-2x}{3}\text{, }x\leq -\frac{3}{4}\text{ or }x\geq 0$$