To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-9x^{2}+ax+bx-1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-6$$ $$ab=-9\left(-1\right)=9$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $9$.
$$-1,-9$$ $$-3,-3$$
Calculate the sum for each pair.
$$-1-9=-10$$ $$-3-3=-6$$
The solution is the pair that gives sum $-6$.
$$a=-3$$ $$b=-3$$
Rewrite $-9x^{2}-6x-1$ as $\left(-9x^{2}-3x\right)+\left(-3x-1\right)$.
$$\left(-9x^{2}-3x\right)+\left(-3x-1\right)$$
Factor out $3x$ in $-9x^{2}-3x$.
$$3x\left(-3x-1\right)-3x-1$$
Factor out common term $-3x-1$ by using distributive property.
$$\left(-3x-1\right)\left(3x+1\right)$$
To find equation solutions, solve $-3x-1=0$ and $3x+1=0$.
$$x=-\frac{1}{3}$$ $$x=-\frac{1}{3}$$
Steps Using the Quadratic Formula
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$-9x^{2}-6x-1=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-9$ for $a$, $-6$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Reduce the fraction $\frac{6}{-18}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{1}{3}$$
Steps for Completing the Square
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
$$-9x^{2}-6x-1=0$$
Add $1$ to both sides of the equation.
$$-9x^{2}-6x-1-\left(-1\right)=-\left(-1\right)$$
Subtracting $-1$ from itself leaves $0$.
$$-9x^{2}-6x=-\left(-1\right)$$
Subtract $-1$ from $0$.
$$-9x^{2}-6x=1$$
Divide both sides by $-9$.
$$\frac{-9x^{2}-6x}{-9}=\frac{1}{-9}$$
Dividing by $-9$ undoes the multiplication by $-9$.
Reduce the fraction $\frac{-6}{-9}$ to lowest terms by extracting and canceling out $3$.
$$x^{2}+\frac{2}{3}x=\frac{1}{-9}$$
Divide $1$ by $-9$.
$$x^{2}+\frac{2}{3}x=-\frac{1}{9}$$
Divide $\frac{2}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{3}$. Then add the square of $\frac{1}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
Square $\frac{1}{3}$ by squaring both the numerator and the denominator of the fraction.
$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{-1+1}{9}$$
Add $-\frac{1}{9}$ to $\frac{1}{9}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
$$x^{2}+\frac{2}{3}x+\frac{1}{9}=0$$
Factor $x^{2}+\frac{2}{3}x+\frac{1}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x+\frac{1}{3}\right)^{2}=0$$
Take the square root of both sides of the equation.
