Solve x^2-6x=-x^2+12x

$$x ^ { 2 } - 6 x = - x ^ { 2 } + 12 x$$

Solve for x

$$x=9$$

$$x=0$$

Steps Using Factoring

Add x^{2} to both sides.

$$x^{2}-6x+x^{2}=12x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-6x=12x$$

Subtract 12x from both sides.

$$2x^{2}-6x-12x=0$$

Combine -6x and -12x to get -18x.

$$2x^{2}-18x=0$$

Factor out x.

$$x\left(2x-18\right)=0$$

To find equation solutions, solve x=0 and 2x-18=0.

$$x=0 x=9$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$x^{2}-6x+x^{2}=12x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-6x=12x$$

Subtract 12x from both sides.

$$2x^{2}-6x-12x=0$$

Combine -6x and -12x to get -18x.

$$2x^{2}-18x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -18 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}}}{2\times 2}$$

Take the square root of \left(-18\right)^{2}.

$$x=\frac{-\left(-18\right)±18}{2\times 2}$$

The opposite of -18 is 18.

$$x=\frac{18±18}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{18±18}{4}$$

Now solve the equation x=\frac{18±18}{4} when ± is plus. Add 18 to 18.

$$x=\frac{36}{4}$$

Divide 36 by 4.

$$x=9$$

Now solve the equation x=\frac{18±18}{4} when ± is minus. Subtract 18 from 18.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=9 x=0$$

Steps for Completing the Square

Add x^{2} to both sides.

$$x^{2}-6x+x^{2}=12x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-6x=12x$$

Subtract 12x from both sides.

$$2x^{2}-6x-12x=0$$

Combine -6x and -12x to get -18x.

$$2x^{2}-18x=0$$

Divide both sides by 2.

$$\frac{2x^{2}-18x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{18}{2}\right)x=\frac{0}{2}$$

Divide -18 by 2.

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

Divide 0 by 2.

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

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.

$$x=9 x=0$$