Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-3\right)^{2}$.
$$x^{2}-6x+9=9$$
Subtract $9$ from both sides.
$$x^{2}-6x+9-9=0$$
Subtract $9$ from $9$ to get $0$.
$$x^{2}-6x=0$$
Factor out $x$.
$$x\left(x-6\right)=0$$
To find equation solutions, solve $x=0$ and $x-6=0$.
$$x=0$$ $$x=6$$
Steps Using the Quadratic Formula
Divide both sides by $3$.
$$\left(x-3\right)^{2}=\frac{27}{3}$$
Divide $27$ by $3$ to get $9$.
$$\left(x-3\right)^{2}=9$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-3\right)^{2}$.
$$x^{2}-6x+9=9$$
Subtract $9$ from both sides.
$$x^{2}-6x+9-9=0$$
Subtract $9$ from $9$ to get $0$.
$$x^{2}-6x=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-6$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
