Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-1\right)^{2}$.
$$x^{2}-2x+1=1-2x$$
Add $2x$ to both sides.
$$x^{2}-2x+1+2x=1$$
Combine $-2x$ and $2x$ to get $0$.
$$x^{2}+1=1$$
Subtract $1$ from both sides.
$$x^{2}=1-1$$
Subtract $1$ from $1$ to get $0$.
$$x^{2}=0$$
Take the square root of both sides of the equation.
$$x=0$$ $$x=0$$
The equation is now solved. Solutions are the same.
$$x=0$$
Steps Using the Quadratic Formula
Multiply both sides of the equation by $2$.
$$\left(x-1\right)^{2}=1-2x$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-1\right)^{2}$.
$$x^{2}-2x+1=1-2x$$
Subtract $1$ from both sides.
$$x^{2}-2x+1-1=-2x$$
Subtract $1$ from $1$ to get $0$.
$$x^{2}-2x=-2x$$
Add $2x$ to both sides.
$$x^{2}-2x+2x=0$$
Combine $-2x$ and $2x$ to get $0$.
$$x^{2}=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
