Solve left(x-4right)^2-4=0

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

Solve for x

$$x=6$$

$$x=2$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-4=0$$

Subtract 4 from 16 to get 12.

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

To solve the equation, factor x^{2}-8x+12 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-6\right)\left(x-2\right)$$

To find equation solutions, solve x-6=0 and x-2=0.

$$x=6 x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-4=0$$

Subtract 4 from 16 to get 12.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 12=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite x^{2}-8x+12 as \left(x^{2}-6x\right)+\left(-2x+12\right).

$$\left(x^{2}-6x\right)+\left(-2x+12\right)$$

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

$$x\left(x-6\right)-2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-2\right)$$

To find equation solutions, solve x-6=0 and x-2=0.

$$x=6 x=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-4=0$$

Subtract 4 from 16 to get 12.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 12}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-\left(-8\right)±\sqrt{64-48}}{2}$$

Add 64 to -48.

$$x=\frac{-\left(-8\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-\left(-8\right)±4}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±4}{2}$$

Now solve the equation x=\frac{8±4}{2} when ± is plus. Add 8 to 4.

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

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{8±4}{2} when ± is minus. Subtract 4 from 8.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=6 x=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-4=0$$

Subtract 4 from 16 to get 12.

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

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-8x=-12$$

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.

$$x^{2}-8x+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-12+16$$

Add -12 to 16.

$$x^{2}-8x+16=4$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=2 x-4=-2$$

Add 4 to both sides of the equation.

$$x=6 x=2$$