Solve 6x^2-3x=x

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

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=0$$

Steps Using Factoring

Subtract x from both sides.

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

Combine -3x and -x to get -4x.

$$6x^{2}-4x=0$$

Factor out x.

$$x\left(6x-4\right)=0$$

To find equation solutions, solve x=0 and 6x-4=0.

$$x=0 x=\frac{2}{3}$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Combine -3x and -x to get -4x.

$$6x^{2}-4x=0$$

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

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

Take the square root of \left(-4\right)^{2}.

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

The opposite of -4 is 4.

$$x=\frac{4±4}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{4±4}{12}$$

Now solve the equation x=\frac{4±4}{12} when ± is plus. Add 4 to 4.

$$x=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{4±4}{12} when ± is minus. Subtract 4 from 4.

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

Divide 0 by 12.

$$x=0$$

The equation is now solved.

$$x=\frac{2}{3} x=0$$

Steps for Completing the Square

Subtract x from both sides.

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

Combine -3x and -x to get -4x.

$$6x^{2}-4x=0$$

Divide both sides by 6.

$$\frac{6x^{2}-4x}{6}=\frac{0}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{4}{6}\right)x=\frac{0}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

Divide 0 by 6.

$$x^{2}-\frac{2}{3}x=0$$

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=\frac{2}{3} x=0$$