Solve {left(2(x-2)/1right)}^2-2*16/1=16

$${ \left( \frac{ 2(x-2) }{ 1 } \right) }^{ 2 } -2 \times \frac{ 16 }{ 1 } =16$$

Solve for x

$$x=2\sqrt{3}+2\approx 5.464101615$$

$$x=2-2\sqrt{3}\approx -1.464101615$$

Steps Using the Quadratic Formula

Anything divided by one gives itself.

$$\left(2\left(x-2\right)\right)^{2}-2\times \frac{16}{1}=16$$

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

$$\left(2x-4\right)^{2}-2\times \frac{16}{1}=16$$

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

$$4x^{2}-16x+16-2\times \frac{16}{1}=16$$

Anything divided by one gives itself.

$$4x^{2}-16x+16-2\times 16=16$$

Multiply 2 and 16 to get 32.

$$4x^{2}-16x+16-32=16$$

Subtract 32 from 16 to get -16.

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

Subtract 16 from both sides.

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

Subtract 16 from -16 to get -32.

$$4x^{2}-16x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -16 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 4\left(-32\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-16\right)±\sqrt{256-16\left(-32\right)}}{2\times 4}$$

Multiply -16 times -32.

$$x=\frac{-\left(-16\right)±\sqrt{256+512}}{2\times 4}$$

Add 256 to 512.

$$x=\frac{-\left(-16\right)±\sqrt{768}}{2\times 4}$$

Take the square root of 768.

$$x=\frac{-\left(-16\right)±16\sqrt{3}}{2\times 4}$$

The opposite of -16 is 16.

$$x=\frac{16±16\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{16±16\sqrt{3}}{8}$$

Now solve the equation x=\frac{16±16\sqrt{3}}{8} when ± is plus. Add 16 to 16\sqrt{3}.

$$x=\frac{16\sqrt{3}+16}{8}$$

Divide 16+16\sqrt{3} by 8.

$$x=2\sqrt{3}+2$$

Now solve the equation x=\frac{16±16\sqrt{3}}{8} when ± is minus. Subtract 16\sqrt{3} from 16.

$$x=\frac{16-16\sqrt{3}}{8}$$

Divide 16-16\sqrt{3} by 8.

$$x=2-2\sqrt{3}$$

The equation is now solved.

$$x=2\sqrt{3}+2 x=2-2\sqrt{3}$$

Steps for Completing the Square

Anything divided by one gives itself.

$$\left(2\left(x-2\right)\right)^{2}-2\times \frac{16}{1}=16$$

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

$$\left(2x-4\right)^{2}-2\times \frac{16}{1}=16$$

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

$$4x^{2}-16x+16-2\times \frac{16}{1}=16$$

Anything divided by one gives itself.

$$4x^{2}-16x+16-2\times 16=16$$

Multiply 2 and 16 to get 32.

$$4x^{2}-16x+16-32=16$$

Subtract 32 from 16 to get -16.

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

Add 16 to both sides.

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

Add 16 and 16 to get 32.

$$4x^{2}-16x=32$$

Divide both sides by 4.

$$\frac{4x^{2}-16x}{4}=\frac{32}{4}$$

Dividing by 4 undoes the multiplication by 4.

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

Divide -16 by 4.

$$x^{2}-4x=\frac{32}{4}$$

Divide 32 by 4.

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

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.

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

Square -2.

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

Add 8 to 4.

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

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{12}$$

Simplify.

$$x-2=2\sqrt{3} x-2=-2\sqrt{3}$$

Add 2 to both sides of the equation.

$$x=2\sqrt{3}+2 x=2-2\sqrt{3}$$