To find equation solutions, solve $x=0$ and $3x-9=0$.
$$x=0$$ $$x=3$$
Steps Using the Quadratic Formula
Subtract $9x$ from both sides.
$$3x^{2}-9x=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $-9$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $x=\frac{9±9}{6}$ when $±$ is plus. Add $9$ to $9$.
$$x=\frac{18}{6}$$
Divide $18$ by $6$.
$$x=3$$
Now solve the equation $x=\frac{9±9}{6}$ when $±$ is minus. Subtract $9$ from $9$.
$$x=\frac{0}{6}$$
Divide $0$ by $6$.
$$x=0$$
The equation is now solved.
$$x=3$$ $$x=0$$
Steps for Completing the Square
Subtract $9x$ from both sides.
$$3x^{2}-9x=0$$
Divide both sides by $3$.
$$\frac{3x^{2}-9x}{3}=\frac{0}{3}$$
Dividing by $3$ undoes the multiplication by $3$.
$$x^{2}+\left(-\frac{9}{3}\right)x=\frac{0}{3}$$
Divide $-9$ by $3$.
$$x^{2}-3x=\frac{0}{3}$$
Divide $0$ by $3$.
$$x^{2}-3x=0$$
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.
