Solve (x-2)^2-4*(x-2)-4+x^2=0

$$( x - 2 ) ^ { 2 } - 4 \cdot ( x - 2 ) - 4 + x ^ { 2 } = 0$$

Solve for x

$$x=2$$

Steps Using Factoring By Grouping

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

$$x^{2}-4x+4-4\left(x-2\right)-4+x^{2}=0$$

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

$$x^{2}-4x+4-4x+8-4+x^{2}=0$$

Combine -4x and -4x to get -8x.

$$x^{2}-8x+4+8-4+x^{2}=0$$

Add 4 and 8 to get 12.

$$x^{2}-8x+12-4+x^{2}=0$$

Subtract 4 from 12 to get 8.

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

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

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

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

Rewrite x^{2}-4x+4 as \left(x^{2}-2x\right)+\left(-2x+4\right).

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

Factor out x in the first and -2 in the second group.

$$x\left(x-2\right)-2\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x-2\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x-2=0.

$$x=2$$

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}.

$$x^{2}-4x+4-4\left(x-2\right)-4+x^{2}=0$$

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

$$x^{2}-4x+4-4x+8-4+x^{2}=0$$

Combine -4x and -4x to get -8x.

$$x^{2}-8x+4+8-4+x^{2}=0$$

Add 4 and 8 to get 12.

$$x^{2}-8x+12-4+x^{2}=0$$

Subtract 4 from 12 to get 8.

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

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

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

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

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\times 8}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 8}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 8}}{2\times 2}$$

Multiply -8 times 8.

$$x=\frac{-\left(-8\right)±\sqrt{64-64}}{2\times 2}$$

Add 64 to -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

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

The opposite of -8 is 8.

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

Multiply 2 times 2.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

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}.

$$x^{2}-4x+4-4\left(x-2\right)-4+x^{2}=0$$

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

$$x^{2}-4x+4-4x+8-4+x^{2}=0$$

Combine -4x and -4x to get -8x.

$$x^{2}-8x+4+8-4+x^{2}=0$$

Add 4 and 8 to get 12.

$$x^{2}-8x+12-4+x^{2}=0$$

Subtract 4 from 12 to get 8.

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

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

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

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -8 by 2.

$$x^{2}-4x=-\frac{8}{2}$$

Divide -8 by 2.

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

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.

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

Square -2.

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

Add -4 to 4.

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

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-2=0 x-2=0$$

Add 2 to both sides of the equation.

$$x=2 x=2$$

The equation is now solved. Solutions are the same.

$$x=2$$