Solve 0=3x^2-3x-6

$$0 = 3 x ^ { 2 } - 3 x - 6$$

Solve for x

$$x=-1$$

$$x=2$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-3x-6=0$$

Divide both sides by 3.

$$x^{2}-x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-2\right)=-2$$

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.

$$a=-2 b=1$$

Rewrite x^{2}-x-2 as \left(x^{2}-2x\right)+\left(x-2\right).

$$\left(x^{2}-2x\right)+\left(x-2\right)$$

Factor out x in x^{2}-2x.

$$x\left(x-2\right)+x-2$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+1\right)$$

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-3x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -3 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-6\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-3\right)±\sqrt{9-12\left(-6\right)}}{2\times 3}$$

Multiply -12 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\times 3}$$

Add 9 to 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\times 3}$$

Take the square root of 81.

$$x=\frac{-\left(-3\right)±9}{2\times 3}$$

The opposite of -3 is 3.

$$x=\frac{3±9}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{3±9}{6}$$

Now solve the equation x=\frac{3±9}{6} when ± is plus. Add 3 to 9.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{3±9}{6} when ± is minus. Subtract 9 from 3.

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=2 x=-1$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-3x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$3x^{2}-3x=6$$

Divide both sides by 3.

$$\frac{3x^{2}-3x}{3}=\frac{6}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{3}{3}\right)x=\frac{6}{3}$$

Divide -3 by 3.

$$x^{2}-x=\frac{6}{3}$$

Divide 6 by 3.

$$x^{2}-x=2$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=2+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{3}{2} x-\frac{1}{2}=-\frac{3}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$