Solve (q-r)^2+(r-p)^2+(x-q)^2=(p-x)^2

$$( q - r ) ^ { 2 } + ( r - p ) ^ { 2 } + ( x - q ) ^ { 2 } = ( p - x ) ^ { 2 }$$

Solve for p (complex solution)

$$\left\{\begin{matrix}p=-\frac{r^{2}+q^{2}-qr-qx}{x-r}\text{, }&r\neq x\\p\in \mathrm{C}\text{, }&r=x\text{ and }q=r\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$q^{2}-2qr+r^{2}+\left(r-p\right)^{2}+\left(x-q\right)^{2}=\left(p-x\right)^{2}$$

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

$$q^{2}-2qr+r^{2}+r^{2}-2rp+p^{2}+\left(x-q\right)^{2}=\left(p-x\right)^{2}$$

Combine r^{2} and r^{2} to get 2r^{2}.

$$q^{2}-2qr+2r^{2}-2rp+p^{2}+\left(x-q\right)^{2}=\left(p-x\right)^{2}$$

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

$$q^{2}-2qr+2r^{2}-2rp+p^{2}+x^{2}-2xq+q^{2}=\left(p-x\right)^{2}$$

Combine q^{2} and q^{2} to get 2q^{2}.

$$2q^{2}-2qr+2r^{2}-2rp+p^{2}+x^{2}-2xq=\left(p-x\right)^{2}$$

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

$$2q^{2}-2qr+2r^{2}-2rp+p^{2}+x^{2}-2xq=p^{2}-2px+x^{2}$$

Subtract p^{2} from both sides.

$$2q^{2}-2qr+2r^{2}-2rp+p^{2}+x^{2}-2xq-p^{2}=-2px+x^{2}$$

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

$$2q^{2}-2qr+2r^{2}-2rp+x^{2}-2xq=-2px+x^{2}$$

Add 2px to both sides.

$$2q^{2}-2qr+2r^{2}-2rp+x^{2}-2xq+2px=x^{2}$$

Subtract 2q^{2} from both sides.

$$-2qr+2r^{2}-2rp+x^{2}-2xq+2px=x^{2}-2q^{2}$$

Add 2qr to both sides.

$$2r^{2}-2rp+x^{2}-2xq+2px=x^{2}-2q^{2}+2qr$$

Subtract 2r^{2} from both sides.

$$-2rp+x^{2}-2xq+2px=x^{2}-2q^{2}+2qr-2r^{2}$$

Subtract x^{2} from both sides.

$$-2rp-2xq+2px=x^{2}-2q^{2}+2qr-2r^{2}-x^{2}$$

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

$$-2rp-2xq+2px=-2q^{2}+2qr-2r^{2}$$

Add 2xq to both sides.

$$-2rp+2px=-2q^{2}+2qr-2r^{2}+2xq$$

Combine all terms containing p.

$$\left(-2r+2x\right)p=-2q^{2}+2qr-2r^{2}+2xq$$

The equation is in standard form.

$$\left(2x-2r\right)p=2qx+2qr-2q^{2}-2r^{2}$$

Divide both sides by 2x-2r.

$$\frac{\left(2x-2r\right)p}{2x-2r}=\frac{2qx+2qr-2q^{2}-2r^{2}}{2x-2r}$$

Dividing by 2x-2r undoes the multiplication by 2x-2r.

$$p=\frac{2qx+2qr-2q^{2}-2r^{2}}{2x-2r}$$

Divide -2q^{2}+2qr-2r^{2}+2xq by 2x-2r.

$$p=\frac{qx+qr-q^{2}-r^{2}}{x-r}$$

Solve for p

$$\left\{\begin{matrix}p=-\frac{r^{2}+q^{2}-qr-qx}{x-r}\text{, }&r\neq x\\p\in \mathrm{R}\text{, }&r=x\text{ and }q=r\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$q^{2}-2qr+r^{2}+\left(r-p\right)^{2}+\left(x-q\right)^{2}=\left(p-x\right)^{2}$$

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

$$q^{2}-2qr+r^{2}+r^{2}-2rp+p^{2}+\left(x-q\right)^{2}=\left(p-x\right)^{2}$$

Combine r^{2} and r^{2} to get 2r^{2}.

$$q^{2}-2qr+2r^{2}-2rp+p^{2}+\left(x-q\right)^{2}=\left(p-x\right)^{2}$$

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

$$q^{2}-2qr+2r^{2}-2rp+p^{2}+x^{2}-2xq+q^{2}=\left(p-x\right)^{2}$$

Combine q^{2} and q^{2} to get 2q^{2}.

$$2q^{2}-2qr+2r^{2}-2rp+p^{2}+x^{2}-2xq=\left(p-x\right)^{2}$$

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

$$2q^{2}-2qr+2r^{2}-2rp+p^{2}+x^{2}-2xq=p^{2}-2px+x^{2}$$

Subtract p^{2} from both sides.

$$2q^{2}-2qr+2r^{2}-2rp+p^{2}+x^{2}-2xq-p^{2}=-2px+x^{2}$$

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

$$2q^{2}-2qr+2r^{2}-2rp+x^{2}-2xq=-2px+x^{2}$$

Add 2px to both sides.

$$2q^{2}-2qr+2r^{2}-2rp+x^{2}-2xq+2px=x^{2}$$

Subtract 2q^{2} from both sides.

$$-2qr+2r^{2}-2rp+x^{2}-2xq+2px=x^{2}-2q^{2}$$

Add 2qr to both sides.

$$2r^{2}-2rp+x^{2}-2xq+2px=x^{2}-2q^{2}+2qr$$

Subtract 2r^{2} from both sides.

$$-2rp+x^{2}-2xq+2px=x^{2}-2q^{2}+2qr-2r^{2}$$

Subtract x^{2} from both sides.

$$-2rp-2xq+2px=x^{2}-2q^{2}+2qr-2r^{2}-x^{2}$$

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

$$-2rp-2xq+2px=-2q^{2}+2qr-2r^{2}$$

Add 2xq to both sides.

$$-2rp+2px=-2q^{2}+2qr-2r^{2}+2xq$$

Combine all terms containing p.

$$\left(-2r+2x\right)p=-2q^{2}+2qr-2r^{2}+2xq$$

The equation is in standard form.

$$\left(2x-2r\right)p=2qx+2qr-2q^{2}-2r^{2}$$

Divide both sides by 2x-2r.

$$\frac{\left(2x-2r\right)p}{2x-2r}=\frac{2qx+2qr-2q^{2}-2r^{2}}{2x-2r}$$

Dividing by 2x-2r undoes the multiplication by 2x-2r.

$$p=\frac{2qx+2qr-2q^{2}-2r^{2}}{2x-2r}$$

Divide -2q^{2}+2qr-2r^{2}+2xq by 2x-2r.

$$p=\frac{qx+qr-q^{2}-r^{2}}{x-r}$$

Solve for q (complex solution)

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

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

Solve for q

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

$$q=\frac{-\sqrt{\left(x-r\right)\left(x+3r-4p\right)}+r+x}{2}\text{, }r=x\text{ or }\left(p\neq x\text{ and }r\leq \frac{2\sqrt{x^{2}-2px+p^{2}}+x+2p}{3}\text{ and }r\geq \frac{-2\sqrt{x^{2}-2px+p^{2}}+x+2p}{3}\right)$$