To find equation solutions, solve $x=0$ and $4x+12=0$.
$$x=0$$ $$x=-3$$
Steps Using the Quadratic Formula
Add $12x$ to both sides.
$$4x^{2}+12x=0$$
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, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-12±\sqrt{12^{2}}}{2\times 4}$$
Take the square root of $12^{2}$.
$$x=\frac{-12±12}{2\times 4}$$
Multiply $2$ times $4$.
$$x=\frac{-12±12}{8}$$
Now solve the equation $x=\frac{-12±12}{8}$ when $±$ is plus. Add $-12$ to $12$.
$$x=\frac{0}{8}$$
Divide $0$ by $8$.
$$x=0$$
Now solve the equation $x=\frac{-12±12}{8}$ when $±$ is minus. Subtract $12$ from $-12$.
$$x=-\frac{24}{8}$$
Divide $-24$ by $8$.
$$x=-3$$
The equation is now solved.
$$x=0$$ $$x=-3$$
Steps for Completing the Square
Add $12x$ to both sides.
$$4x^{2}+12x=0$$
Divide both sides by $4$.
$$\frac{4x^{2}+12x}{4}=\frac{0}{4}$$
Dividing by $4$ undoes the multiplication by $4$.
$$x^{2}+\frac{12}{4}x=\frac{0}{4}$$
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.
