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

$$2 = ( 2 - x ) ^ { 2 } - 2$$

Solve for x

$$x=0$$

$$x=4$$

Steps Using Factoring

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

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

Subtract 2 from 4 to get 2.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

Factor out x.

$$x\left(-4+x\right)=0$$

To find equation solutions, solve x=0 and -4+x=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

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

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

Subtract 2 from 4 to get 2.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Take the square root of \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2}$$

Now solve the equation x=\frac{4±4}{2} when ± is plus. Add 4 to 4.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{4±4}{2} when ± is minus. Subtract 4 from 4.

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

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=4 x=0$$

Steps for Completing the Square

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

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

Subtract 2 from 4 to get 2.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides.

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

Subtract 2 from 2 to get 0.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

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

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

Square -2.

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

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=2 x-2=-2$$

Add 2 to both sides of the equation.

$$x=4 x=0$$