Solve (x-2)^2=36

$$( x - 2 ) ^ { 2 } = 36$$

Solve for x

$$x=8$$

$$x=-4$$

Steps Using Factoring

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

$$x^{2}-4x+4=36$$

Subtract 36 from both sides.

$$x^{2}-4x+4-36=0$$

Subtract 36 from 4 to get -32.

$$x^{2}-4x-32=0$$

To solve the equation, factor x^{2}-4x-32 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.

$$a+b=-4 ab=-32$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -32.

$$1,-32 2,-16 4,-8$$

Calculate the sum for each pair.

$$1-32=-31 2-16=-14 4-8=-4$$

The solution is the pair that gives sum -4.

$$a=-8 b=4$$

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

$$\left(x-8\right)\left(x+4\right)$$

To find equation solutions, solve x-8=0 and x+4=0.

$$x=8 x=-4$$

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}.

$$x^{2}-4x+4=36$$

Subtract 36 from both sides.

$$x^{2}-4x+4-36=0$$

Subtract 36 from 4 to get -32.

$$x^{2}-4x-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-32. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-32\right)=-32$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -32.

$$1,-32 2,-16 4,-8$$

Calculate the sum for each pair.

$$1-32=-31 2-16=-14 4-8=-4$$

The solution is the pair that gives sum -4.

$$a=-8 b=4$$

Rewrite x^{2}-4x-32 as \left(x^{2}-8x\right)+\left(4x-32\right).

$$\left(x^{2}-8x\right)+\left(4x-32\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x-8\right)+4\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+4\right)$$

To find equation solutions, solve x-8=0 and x+4=0.

$$x=8 x=-4$$

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}.

$$x^{2}-4x+4=36$$

Subtract 36 from both sides.

$$x^{2}-4x+4-36=0$$

Subtract 36 from 4 to get -32.

$$x^{2}-4x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-32\right)}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-32\right)}}{2}$$

Multiply -4 times -32.

$$x=\frac{-\left(-4\right)±\sqrt{16+128}}{2}$$

Add 16 to 128.

$$x=\frac{-\left(-4\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-\left(-4\right)±12}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±12}{2}$$

Now solve the equation x=\frac{4±12}{2} when ± is plus. Add 4 to 12.

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

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{4±12}{2} when ± is minus. Subtract 12 from 4.

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

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=8 x=-4$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{36}$$

Simplify.

$$x-2=6 x-2=-6$$

Add 2 to both sides of the equation.

$$x=8 x=-4$$