Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-1\right)^{2}$.
$$4x^{2}-4x+1=\left(x+1\right)^{2}$$
Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+1\right)^{2}$.
$$4x^{2}-4x+1=x^{2}+2x+1$$
Subtract $x^{2}$ from both sides.
$$4x^{2}-4x+1-x^{2}=2x+1$$
Combine $4x^{2}$ and $-x^{2}$ to get $3x^{2}$.
$$3x^{2}-4x+1=2x+1$$
Subtract $2x$ from both sides.
$$3x^{2}-4x+1-2x=1$$
Combine $-4x$ and $-2x$ to get $-6x$.
$$3x^{2}-6x+1=1$$
Subtract $1$ from both sides.
$$3x^{2}-6x+1-1=0$$
Subtract $1$ from $1$ to get $0$.
$$3x^{2}-6x=0$$
Factor out $x$.
$$x\left(3x-6\right)=0$$
To find equation solutions, solve $x=0$ and $3x-6=0$.
$$x=0$$ $$x=2$$
Steps Using the Quadratic Formula
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-1\right)^{2}$.
$$4x^{2}-4x+1=\left(x+1\right)^{2}$$
Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+1\right)^{2}$.
$$4x^{2}-4x+1=x^{2}+2x+1$$
Subtract $x^{2}$ from both sides.
$$4x^{2}-4x+1-x^{2}=2x+1$$
Combine $4x^{2}$ and $-x^{2}$ to get $3x^{2}$.
$$3x^{2}-4x+1=2x+1$$
Subtract $2x$ from both sides.
$$3x^{2}-4x+1-2x=1$$
Combine $-4x$ and $-2x$ to get $-6x$.
$$3x^{2}-6x+1=1$$
Subtract $1$ from both sides.
$$3x^{2}-6x+1-1=0$$
Subtract $1$ from $1$ to get $0$.
$$3x^{2}-6x=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $-6$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{6±6}{6}$ when $±$ is plus. Add $6$ to $6$.
$$x=\frac{12}{6}$$
Divide $12$ by $6$.
$$x=2$$
Now solve the equation $x=\frac{6±6}{6}$ when $±$ is minus. Subtract $6$ from $6$.
$$x=\frac{0}{6}$$
Divide $0$ by $6$.
$$x=0$$
The equation is now solved.
$$x=2$$ $$x=0$$
Steps for Completing the Square
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-1\right)^{2}$.
$$4x^{2}-4x+1=\left(x+1\right)^{2}$$
Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+1\right)^{2}$.
$$4x^{2}-4x+1=x^{2}+2x+1$$
Subtract $x^{2}$ from both sides.
$$4x^{2}-4x+1-x^{2}=2x+1$$
Combine $4x^{2}$ and $-x^{2}$ to get $3x^{2}$.
$$3x^{2}-4x+1=2x+1$$
Subtract $2x$ from both sides.
$$3x^{2}-4x+1-2x=1$$
Combine $-4x$ and $-2x$ to get $-6x$.
$$3x^{2}-6x+1=1$$
Subtract $1$ from both sides.
$$3x^{2}-6x=1-1$$
Subtract $1$ from $1$ to get $0$.
$$3x^{2}-6x=0$$
Divide both sides by $3$.
$$\frac{3x^{2}-6x}{3}=\frac{0}{3}$$
Dividing by $3$ undoes the multiplication by $3$.
$$x^{2}+\left(-\frac{6}{3}\right)x=\frac{0}{3}$$
Divide $-6$ by $3$.
$$x^{2}-2x=\frac{0}{3}$$
Divide $0$ by $3$.
$$x^{2}-2x=0$$
Divide $-2$, the coefficient of the $x$ term, by $2$ to get $-1$. Then add the square of $-1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}-2x+1=1$$
Factor $x^{2}-2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x-1\right)^{2}=1$$
Take the square root of both sides of the equation.
