Solve left(x-2right)^2=2x+4

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

Solve for x

$$x=6$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$x^{2}-4x+4=2x+4$$

Subtract 2x from both sides.

$$x^{2}-4x+4-2x=4$$

Combine -4x and -2x to get -6x.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Factor out x.

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

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

$$x=0 x=6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$x^{2}-4x+4=2x+4$$

Subtract 2x from both sides.

$$x^{2}-4x+4-2x=4$$

Combine -4x and -2x to get -6x.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-6\right)^{2}.

$$x=\frac{-\left(-6\right)±6}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±6}{2}$$

Now solve the equation x=\frac{6±6}{2} when ± is plus. Add 6 to 6.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{6±6}{2} when ± is minus. Subtract 6 from 6.

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

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=6 x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$x^{2}-4x+4=2x+4$$

Subtract 2x from both sides.

$$x^{2}-4x+4-2x=4$$

Combine -4x and -2x to get -6x.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=9$$

Factor x^{2}-6x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=3 x-3=-3$$

Add 3 to both sides of the equation.

$$x=6 x=0$$