Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $3x$, the least common multiple of $3,x$.
Combine $3x^{2}$ and $x^{2}\times 3$ to get $6x^{2}$.
$$6x^{2}+3x\times 9=3x\times 4-3x+3x\times 3$$
Multiply $3$ and $9$ to get $27$.
$$6x^{2}+27x=3x\times 4-3x+3x\times 3$$
Multiply $3$ and $4$ to get $12$.
$$6x^{2}+27x=12x-3x+3x\times 3$$
Combine $12x$ and $-3x$ to get $9x$.
$$6x^{2}+27x=9x+3x\times 3$$
Multiply $3$ and $3$ to get $9$.
$$6x^{2}+27x=9x+9x$$
Combine $9x$ and $9x$ to get $18x$.
$$6x^{2}+27x=18x$$
Subtract $18x$ from both sides.
$$6x^{2}+27x-18x=0$$
Combine $27x$ and $-18x$ to get $9x$.
$$6x^{2}+9x=0$$
Factor out $x$.
$$x\left(6x+9\right)=0$$
To find equation solutions, solve $x=0$ and $6x+9=0$.
$$x=0$$ $$x=-\frac{3}{2}$$
Variable $x$ cannot be equal to $0$.
$$x=-\frac{3}{2}$$
Steps Using the Quadratic Formula
Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $3x$, the least common multiple of $3,x$.
Combine $3x^{2}$ and $x^{2}\times 3$ to get $6x^{2}$.
$$6x^{2}+3x\times 9=3x\times 4-3x+3x\times 3$$
Multiply $3$ and $9$ to get $27$.
$$6x^{2}+27x=3x\times 4-3x+3x\times 3$$
Multiply $3$ and $4$ to get $12$.
$$6x^{2}+27x=12x-3x+3x\times 3$$
Combine $12x$ and $-3x$ to get $9x$.
$$6x^{2}+27x=9x+3x\times 3$$
Multiply $3$ and $3$ to get $9$.
$$6x^{2}+27x=9x+9x$$
Combine $9x$ and $9x$ to get $18x$.
$$6x^{2}+27x=18x$$
Subtract $18x$ from both sides.
$$6x^{2}+27x-18x=0$$
Combine $27x$ and $-18x$ to get $9x$.
$$6x^{2}+9x=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $6$ for $a$, $9$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{-9±\sqrt{9^{2}}}{2\times 6}$$
Take the square root of $9^{2}$.
$$x=\frac{-9±9}{2\times 6}$$
Multiply $2$ times $6$.
$$x=\frac{-9±9}{12}$$
Now solve the equation $x=\frac{-9±9}{12}$ when $±$ is plus. Add $-9$ to $9$.
$$x=\frac{0}{12}$$
Divide $0$ by $12$.
$$x=0$$
Now solve the equation $x=\frac{-9±9}{12}$ when $±$ is minus. Subtract $9$ from $-9$.
$$x=-\frac{18}{12}$$
Reduce the fraction $\frac{-18}{12}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{3}{2}$$
The equation is now solved.
$$x=0$$ $$x=-\frac{3}{2}$$
Variable $x$ cannot be equal to $0$.
$$x=-\frac{3}{2}$$
Steps for Completing the Square
Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $3x$, the least common multiple of $3,x$.
Combine $3x^{2}$ and $x^{2}\times 3$ to get $6x^{2}$.
$$6x^{2}+3x\times 9=3x\times 4-3x+3x\times 3$$
Multiply $3$ and $9$ to get $27$.
$$6x^{2}+27x=3x\times 4-3x+3x\times 3$$
Multiply $3$ and $4$ to get $12$.
$$6x^{2}+27x=12x-3x+3x\times 3$$
Combine $12x$ and $-3x$ to get $9x$.
$$6x^{2}+27x=9x+3x\times 3$$
Multiply $3$ and $3$ to get $9$.
$$6x^{2}+27x=9x+9x$$
Combine $9x$ and $9x$ to get $18x$.
$$6x^{2}+27x=18x$$
Subtract $18x$ from both sides.
$$6x^{2}+27x-18x=0$$
Combine $27x$ and $-18x$ to get $9x$.
$$6x^{2}+9x=0$$
Divide both sides by $6$.
$$\frac{6x^{2}+9x}{6}=\frac{0}{6}$$
Dividing by $6$ undoes the multiplication by $6$.
$$x^{2}+\frac{9}{6}x=\frac{0}{6}$$
Reduce the fraction $\frac{9}{6}$ to lowest terms by extracting and canceling out $3$.
$$x^{2}+\frac{3}{2}x=\frac{0}{6}$$
Divide $0$ by $6$.
$$x^{2}+\frac{3}{2}x=0$$
Divide $\frac{3}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{3}{4}$. Then add the square of $\frac{3}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
Square $\frac{3}{4}$ by squaring both the numerator and the denominator of the fraction.
$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}$$
Factor $x^{2}+\frac{3}{2}x+\frac{9}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+\frac{3}{4}\right)^{2}=\frac{9}{16}$$
Take the square root of both sides of the equation.
