Solve (x+6)^2-16=0

$$( x + 6 ) ^ { 2 } - 16 = 0$$

Solve for x

$$x=-2$$

$$x=-10$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+6\right)^{2}.

$$x^{2}+12x+36-16=0$$

Subtract 16 from 36 to get 20.

$$x^{2}+12x+20=0$$

To solve the equation, factor x^{2}+12x+20 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.

$$a+b=12 ab=20$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 12.

$$a=2 b=10$$

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

$$\left(x+2\right)\left(x+10\right)$$

To find equation solutions, solve x+2=0 and x+10=0.

$$x=-2 x=-10$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+6\right)^{2}.

$$x^{2}+12x+36-16=0$$

Subtract 16 from 36 to get 20.

$$x^{2}+12x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 20=20$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 12.

$$a=2 b=10$$

Rewrite x^{2}+12x+20 as \left(x^{2}+2x\right)+\left(10x+20\right).

$$\left(x^{2}+2x\right)+\left(10x+20\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(x+2\right)+10\left(x+2\right)$$

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

$$\left(x+2\right)\left(x+10\right)$$

To find equation solutions, solve x+2=0 and x+10=0.

$$x=-2 x=-10$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+6\right)^{2}.

$$x^{2}+12x+36-16=0$$

Subtract 16 from 36 to get 20.

$$x^{2}+12x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 20}}{2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 20}}{2}$$

Multiply -4 times 20.

$$x=\frac{-12±\sqrt{144-80}}{2}$$

Add 144 to -80.

$$x=\frac{-12±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-12±8}{2}$$

Now solve the equation x=\frac{-12±8}{2} when ± is plus. Add -12 to 8.

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

Divide -4 by 2.

$$x=-2$$

Now solve the equation x=\frac{-12±8}{2} when ± is minus. Subtract 8 from -12.

$$x=-\frac{20}{2}$$

Divide -20 by 2.

$$x=-10$$

The equation is now solved.

$$x=-2 x=-10$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+6\right)^{2}.

$$x^{2}+12x+36-16=0$$

Subtract 16 from 36 to get 20.

$$x^{2}+12x+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+12x=-20$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+12x+6^{2}=-20+6^{2}$$

Square 6.

$$x^{2}+12x+36=-20+36$$

Add -20 to 36.

$$x^{2}+12x+36=16$$

Factor x^{2}+12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+6\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{16}$$

Simplify.

$$x+6=4 x+6=-4$$

Subtract 6 from both sides of the equation.

$$x=-2 x=-10$$