Solve 2x^2=x/3

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

Solve for x

$$x=\frac{1}{6}\approx 0.166666667$$

$$x=0$$

Steps Using Factoring

Multiply both sides of the equation by 3.

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

Subtract x from both sides.

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

Factor out x.

$$x\left(6x-1\right)=0$$

To find equation solutions, solve x=0 and 6x-1=0.

$$x=0 x=\frac{1}{6}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

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

Subtract x from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\times 6}$$

Take the square root of 1.

$$x=\frac{-\left(-1\right)±1}{2\times 6}$$

The opposite of -1 is 1.

$$x=\frac{1±1}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{1±1}{12}$$

Now solve the equation x=\frac{1±1}{12} when ± is plus. Add 1 to 1.

$$x=\frac{2}{12}$$

Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{6}$$

Now solve the equation x=\frac{1±1}{12} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{12}$$

Divide 0 by 12.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{6} x=0$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

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

Subtract x from both sides.

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

Divide both sides by 6.

$$\frac{6x^{2}-x}{6}=\frac{0}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{1}{6}x=\frac{0}{6}$$

Divide 0 by 6.

$$x^{2}-\frac{1}{6}x=0$$

Divide -\frac{1}{6}, the coefficient of the x term, by 2 to get -\frac{1}{12}. Then add the square of -\frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{6}x+\left(-\frac{1}{12}\right)^{2}=\left(-\frac{1}{12}\right)^{2}$$

Square -\frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{1}{144}$$

Factor x^{2}-\frac{1}{6}x+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$x-\frac{1}{12}=\frac{1}{12} x-\frac{1}{12}=-\frac{1}{12}$$

Add \frac{1}{12} to both sides of the equation.

$$x=\frac{1}{6} x=0$$