Solve 15x^2-2x=x

$$15 x ^ { 2 } - 2 x = x$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=0$$

Steps Using Factoring

Subtract x from both sides.

$$15x^{2}-2x-x=0$$

Combine -2x and -x to get -3x.

$$15x^{2}-3x=0$$

Factor out x.

$$x\left(15x-3\right)=0$$

To find equation solutions, solve x=0 and 15x-3=0.

$$x=0 x=\frac{1}{5}$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$15x^{2}-2x-x=0$$

Combine -2x and -x to get -3x.

$$15x^{2}-3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 15}$$

Take the square root of \left(-3\right)^{2}.

$$x=\frac{-\left(-3\right)±3}{2\times 15}$$

The opposite of -3 is 3.

$$x=\frac{3±3}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{3±3}{30}$$

Now solve the equation x=\frac{3±3}{30} when ± is plus. Add 3 to 3.

$$x=\frac{6}{30}$$

Reduce the fraction \frac{6}{30} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{3±3}{30} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{30}$$

Divide 0 by 30.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{5} x=0$$

Steps for Completing the Square

Subtract x from both sides.

$$15x^{2}-2x-x=0$$

Combine -2x and -x to get -3x.

$$15x^{2}-3x=0$$

Divide both sides by 15.

$$\frac{15x^{2}-3x}{15}=\frac{0}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}+\left(-\frac{3}{15}\right)x=\frac{0}{15}$$

Reduce the fraction \frac{-3}{15} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{1}{5}x=\frac{0}{15}$$

Divide 0 by 15.

$$x^{2}-\frac{1}{5}x=0$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{1}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{1}{10} x-\frac{1}{10}=-\frac{1}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{1}{5} x=0$$