Solve (x+3)^2=16

$$( x + 3 ) ^ { 2 } = 16$$

Solve for x

$$x=1$$

$$x=-7$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=16$$

Subtract 16 from both sides.

$$x^{2}+6x+9-16=0$$

Subtract 16 from 9 to get -7.

$$x^{2}+6x-7=0$$

To solve the equation, factor x^{2}+6x-7 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.

$$a+b=6 ab=-7$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=7$$

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

$$\left(x-1\right)\left(x+7\right)$$

To find equation solutions, solve x-1=0 and x+7=0.

$$x=1 x=-7$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=16$$

Subtract 16 from both sides.

$$x^{2}+6x+9-16=0$$

Subtract 16 from 9 to get -7.

$$x^{2}+6x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-7\right)=-7$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=7$$

Rewrite x^{2}+6x-7 as \left(x^{2}-x\right)+\left(7x-7\right).

$$\left(x^{2}-x\right)+\left(7x-7\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-1\right)+7\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+7\right)$$

To find equation solutions, solve x-1=0 and x+7=0.

$$x=1 x=-7$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=16$$

Subtract 16 from both sides.

$$x^{2}+6x+9-16=0$$

Subtract 16 from 9 to get -7.

$$x^{2}+6x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-7\right)}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-7\right)}}{2}$$

Multiply -4 times -7.

$$x=\frac{-6±\sqrt{36+28}}{2}$$

Add 36 to 28.

$$x=\frac{-6±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-6±8}{2}$$

Now solve the equation x=\frac{-6±8}{2} when ± is plus. Add -6 to 8.

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

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-6±8}{2} when ± is minus. Subtract 8 from -6.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=1 x=-7$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{16}$$

Simplify.

$$x+3=4 x+3=-4$$

Subtract 3 from both sides of the equation.

$$x=1 x=-7$$