Solve x^2=2x*(x-4)

$$x ^ { 2 } = 2 x \cdot ( x - 4 )$$

Solve for x

$$x=8$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 2x by x-4.

$$x^{2}=2x^{2}-8x$$

Subtract 2x^{2} from both sides.

$$x^{2}-2x^{2}=-8x$$

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

$$-x^{2}=-8x$$

Add 8x to both sides.

$$-x^{2}+8x=0$$

Factor out x.

$$x\left(-x+8\right)=0$$

To find equation solutions, solve x=0 and -x+8=0.

$$x=0 x=8$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by x-4.

$$x^{2}=2x^{2}-8x$$

Subtract 2x^{2} from both sides.

$$x^{2}-2x^{2}=-8x$$

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

$$-x^{2}=-8x$$

Add 8x to both sides.

$$-x^{2}+8x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}}}{2\left(-1\right)}$$

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±8}{-2}$$

Now solve the equation x=\frac{-8±8}{-2} when ± is plus. Add -8 to 8.

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

Divide 0 by -2.

$$x=0$$

Now solve the equation x=\frac{-8±8}{-2} when ± is minus. Subtract 8 from -8.

$$x=-\frac{16}{-2}$$

Divide -16 by -2.

$$x=8$$

The equation is now solved.

$$x=0 x=8$$

Steps for Completing the Square

Use the distributive property to multiply 2x by x-4.

$$x^{2}=2x^{2}-8x$$

Subtract 2x^{2} from both sides.

$$x^{2}-2x^{2}=-8x$$

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

$$-x^{2}=-8x$$

Add 8x to both sides.

$$-x^{2}+8x=0$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=\frac{0}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=\frac{0}{-1}$$

Divide 0 by -1.

$$x^{2}-8x=0$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=16$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-4=4 x-4=-4$$

Add 4 to both sides of the equation.

$$x=8 x=0$$