Solve (a-2)^2=16

$$( a - 2 ) ^ { 2 } = 16$$

Solve for a

$$a=6$$

$$a=-2$$

Steps Using Factoring

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

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

Subtract 16 from both sides.

$$a^{2}-4a+4-16=0$$

Subtract 16 from 4 to get -12.

$$a^{2}-4a-12=0$$

To solve the equation, factor a^{2}-4a-12 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=-12$$

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 -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

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

$$\left(a-6\right)\left(a+2\right)$$

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

$$a=6 a=-2$$

Steps Using Factoring By Grouping

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

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

Subtract 16 from both sides.

$$a^{2}-4a+4-16=0$$

Subtract 16 from 4 to get -12.

$$a^{2}-4a-12=0$$

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

$$a+b=-4 ab=1\left(-12\right)=-12$$

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 -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite a^{2}-4a-12 as \left(a^{2}-6a\right)+\left(2a-12\right).

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

Factor out a in the first and 2 in the second group.

$$a\left(a-6\right)+2\left(a-6\right)$$

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

$$\left(a-6\right)\left(a+2\right)$$

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

$$a=6 a=-2$$

Steps Using the Quadratic Formula

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

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

Subtract 16 from both sides.

$$a^{2}-4a+4-16=0$$

Subtract 16 from 4 to get -12.

$$a^{2}-4a-12=0$$

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

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

Square -4.

$$a=\frac{-\left(-4\right)±\sqrt{16-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$a=\frac{-\left(-4\right)±\sqrt{16+48}}{2}$$

Add 16 to 48.

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

Take the square root of 64.

$$a=\frac{-\left(-4\right)±8}{2}$$

The opposite of -4 is 4.

$$a=\frac{4±8}{2}$$

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

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

Divide 12 by 2.

$$a=6$$

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

$$a=-\frac{4}{2}$$

Divide -4 by 2.

$$a=-2$$

The equation is now solved.

$$a=6 a=-2$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(a-2\right)^{2}}=\sqrt{16}$$

Simplify.

$$a-2=4 a-2=-4$$

Add 2 to both sides of the equation.

$$a=6 a=-2$$