Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $3x$.
$$3x\times 3x+3x\left(-6\right)-27=0$$
Multiply $3$ and $3$ to get $9$.
$$9xx+3x\left(-6\right)-27=0$$
Multiply $x$ and $x$ to get $x^{2}$.
$$9x^{2}+3x\left(-6\right)-27=0$$
Multiply $3$ and $-6$ to get $-18$.
$$9x^{2}-18x-27=0$$
Divide both sides by $9$.
$$x^{2}-2x-3=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-2$$ $$ab=1\left(-3\right)=-3$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
$$a=-3$$ $$b=1$$
Rewrite $x^{2}-2x-3$ as $\left(x^{2}-3x\right)+\left(x-3\right)$.
$$\left(x^{2}-3x\right)+\left(x-3\right)$$
Factor out $x$ in $x^{2}-3x$.
$$x\left(x-3\right)+x-3$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(x+1\right)$$
To find equation solutions, solve $x-3=0$ and $x+1=0$.
$$x=3$$ $$x=-1$$
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 $3x$.
$$3x\times 3x+3x\left(-6\right)-27=0$$
Multiply $3$ and $3$ to get $9$.
$$9xx+3x\left(-6\right)-27=0$$
Multiply $x$ and $x$ to get $x^{2}$.
$$9x^{2}+3x\left(-6\right)-27=0$$
Multiply $3$ and $-6$ to get $-18$.
$$9x^{2}-18x-27=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $9$ for $a$, $-18$ for $b$, and $-27$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Divide $-2$, the coefficient of the $x$ term, by $2$ to get $-1$. Then add the square of $-1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}-2x+1=3+1$$
Add $3$ to $1$.
$$x^{2}-2x+1=4$$
Factor $x^{2}-2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x-1\right)^{2}=4$$
Take the square root of both sides of the equation.
