Solve x^2-6x/3=x

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

Solve for x

$$x=9$$

$$x=0$$

Steps Using Factoring

Divide each term of x^{2}-6x by 3 to get \frac{1}{3}x^{2}-2x.

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

Subtract x from both sides.

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

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

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

Factor out x.

$$x\left(\frac{1}{3}x-3\right)=0$$

To find equation solutions, solve x=0 and \frac{x}{3}-3=0.

$$x=0 x=9$$

Steps Using the Quadratic Formula

Divide each term of x^{2}-6x by 3 to get \frac{1}{3}x^{2}-2x.

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

Subtract x from both sides.

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

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

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

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

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

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

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

The opposite of -3 is 3.

$$x=\frac{3±3}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x=\frac{3±3}{\frac{2}{3}}$$

Now solve the equation x=\frac{3±3}{\frac{2}{3}} when ± is plus. Add 3 to 3.

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

Divide 6 by \frac{2}{3} by multiplying 6 by the reciprocal of \frac{2}{3}.

$$x=9$$

Now solve the equation x=\frac{3±3}{\frac{2}{3}} when ± is minus. Subtract 3 from 3.

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

Divide 0 by \frac{2}{3} by multiplying 0 by the reciprocal of \frac{2}{3}.

$$x=0$$

The equation is now solved.

$$x=9 x=0$$

Steps for Completing the Square

Divide each term of x^{2}-6x by 3 to get \frac{1}{3}x^{2}-2x.

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

Subtract x from both sides.

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

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

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

Multiply both sides by 3.

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

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$x^{2}+\left(-\frac{3}{\frac{1}{3}}\right)x=\frac{0}{\frac{1}{3}}$$

Divide -3 by \frac{1}{3} by multiplying -3 by the reciprocal of \frac{1}{3}.

$$x^{2}-9x=\frac{0}{\frac{1}{3}}$$

Divide 0 by \frac{1}{3} by multiplying 0 by the reciprocal of \frac{1}{3}.

$$x^{2}-9x=0$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=\frac{81}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{9}{2} x-\frac{9}{2}=-\frac{9}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=9 x=0$$