Solve x_{0}x*x-x*x+x=1

$$x _ { 0 } x \cdot x - x \cdot x + x = 1$$

Solve for x_0

$$x_{0}=1-\frac{1}{x}+\frac{1}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply x and x to get x^{2}.

$$x_{0}x^{2}-xx+x=1$$

Multiply x and x to get x^{2}.

$$x_{0}x^{2}-x^{2}+x=1$$

Add x^{2} to both sides.

$$x_{0}x^{2}+x=1+x^{2}$$

Subtract x from both sides.

$$x_{0}x^{2}=1+x^{2}-x$$

The equation is in standard form.

$$x^{2}x_{0}=x^{2}-x+1$$

Divide both sides by x^{2}.

$$\frac{x^{2}x_{0}}{x^{2}}=\frac{x^{2}-x+1}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$x_{0}=\frac{x^{2}-x+1}{x^{2}}$$

Divide 1+x^{2}-x by x^{2}.

$$x_{0}=1-\frac{1}{x}+\frac{1}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{4x_{0}-3}-1}{2\left(x_{0}-1\right)}\text{; }x=-\frac{\sqrt{4x_{0}-3}+1}{2\left(x_{0}-1\right)}\text{, }&x_{0}\neq 1\\x=1\text{, }&x_{0}=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{4x_{0}-3}-1}{2\left(x_{0}-1\right)}\text{; }x=-\frac{\sqrt{4x_{0}-3}+1}{2\left(x_{0}-1\right)}\text{, }&x_{0}\neq 1\text{ and }x_{0}\geq \frac{3}{4}\\x=1\text{, }&x_{0}=1\end{matrix}\right.$$