Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-3\right)^{2}$.
$$4x^{2}-12x+9=9$$
Subtract $9$ from both sides.
$$4x^{2}-12x+9-9=0$$
Subtract $9$ from $9$ to get $0$.
$$4x^{2}-12x=0$$
Factor out $x$.
$$x\left(4x-12\right)=0$$
To find equation solutions, solve $x=0$ and $4x-12=0$.
$$x=0$$ $$x=3$$
Steps Using the Quadratic Formula
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-3\right)^{2}$.
$$4x^{2}-12x+9=9$$
Subtract $9$ from both sides.
$$4x^{2}-12x+9-9=0$$
Subtract $9$ from $9$ to get $0$.
$$4x^{2}-12x=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $4$ for $a$, $-12$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{12±12}{8}$ when $±$ is plus. Add $12$ to $12$.
$$x=\frac{24}{8}$$
Divide $24$ by $8$.
$$x=3$$
Now solve the equation $x=\frac{12±12}{8}$ when $±$ is minus. Subtract $12$ from $12$.
$$x=\frac{0}{8}$$
Divide $0$ by $8$.
$$x=0$$
The equation is now solved.
$$x=3$$ $$x=0$$
Steps for Completing the Square
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-3\right)^{2}$.
$$4x^{2}-12x+9=9$$
Subtract $9$ from both sides.
$$4x^{2}-12x=9-9$$
Subtract $9$ from $9$ to get $0$.
$$4x^{2}-12x=0$$
Divide both sides by $4$.
$$\frac{4x^{2}-12x}{4}=\frac{0}{4}$$
Dividing by $4$ undoes the multiplication by $4$.
$$x^{2}+\left(-\frac{12}{4}\right)x=\frac{0}{4}$$
Divide $-12$ by $4$.
$$x^{2}-3x=\frac{0}{4}$$
Divide $0$ by $4$.
$$x^{2}-3x=0$$
Divide $-3$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{2}$. Then add the square of $-\frac{3}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
