Solve 0=3x^2-2x-1

$$0 = 3 x ^ { 2 } - 2 x - 1$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=1$$

Steps Using Factoring By Grouping

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

$$3x^{2}-2x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-1\right)=-3$$

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 3x^{2}-2x-1 as \left(3x^{2}-3x\right)+\left(x-1\right).

$$\left(3x^{2}-3x\right)+\left(x-1\right)$$

Factor out 3x in 3x^{2}-3x.

$$3x\left(x-1\right)+x-1$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(3x+1\right)$$

To find equation solutions, solve x-1=0 and 3x+1=0.

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

Steps Using the Quadratic Formula

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

$$3x^{2}-2x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-1\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12\left(-1\right)}}{2\times 3}$$

Multiply -12 times -1.

$$x=\frac{-\left(-2\right)±\sqrt{4+12}}{2\times 3}$$

Add 4 to 12.

$$x=\frac{-\left(-2\right)±\sqrt{16}}{2\times 3}$$

Take the square root of 16.

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

The opposite of -2 is 2.

$$x=\frac{2±4}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±4}{6}$$

Now solve the equation x=\frac{2±4}{6} when ± is plus. Add 2 to 4.

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

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{2±4}{6} when ± is minus. Subtract 4 from 2.

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

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$3x^{2}-2x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$3x^{2}-2x=1$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

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.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{1}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{1}{3}+\frac{1}{9}$$

Add \frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{4}{9}$$

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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

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