Solve (n+3)^2=121

$$( n + 3 ) ^ { 2 } = 121$$

Solve for n

$$n=8$$

$$n=-14$$

Steps Using Factoring

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

$$n^{2}+6n+9=121$$

Subtract 121 from both sides.

$$n^{2}+6n+9-121=0$$

Subtract 121 from 9 to get -112.

$$n^{2}+6n-112=0$$

To solve the equation, factor n^{2}+6n-112 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=6 ab=-112$$

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. List all such integer pairs that give product -112.

$$-1,112 -2,56 -4,28 -7,16 -8,14$$

Calculate the sum for each pair.

$$-1+112=111 -2+56=54 -4+28=24 -7+16=9 -8+14=6$$

The solution is the pair that gives sum 6.

$$a=-8 b=14$$

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

$$\left(n-8\right)\left(n+14\right)$$

To find equation solutions, solve n-8=0 and n+14=0.

$$n=8 n=-14$$

Steps Using Factoring By Grouping

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

$$n^{2}+6n+9=121$$

Subtract 121 from both sides.

$$n^{2}+6n+9-121=0$$

Subtract 121 from 9 to get -112.

$$n^{2}+6n-112=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-112. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-112\right)=-112$$

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. List all such integer pairs that give product -112.

$$-1,112 -2,56 -4,28 -7,16 -8,14$$

Calculate the sum for each pair.

$$-1+112=111 -2+56=54 -4+28=24 -7+16=9 -8+14=6$$

The solution is the pair that gives sum 6.

$$a=-8 b=14$$

Rewrite n^{2}+6n-112 as \left(n^{2}-8n\right)+\left(14n-112\right).

$$\left(n^{2}-8n\right)+\left(14n-112\right)$$

Factor out n in the first and 14 in the second group.

$$n\left(n-8\right)+14\left(n-8\right)$$

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

$$\left(n-8\right)\left(n+14\right)$$

To find equation solutions, solve n-8=0 and n+14=0.

$$n=8 n=-14$$

Steps Using the Quadratic Formula

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

$$n^{2}+6n+9=121$$

Subtract 121 from both sides.

$$n^{2}+6n+9-121=0$$

Subtract 121 from 9 to get -112.

$$n^{2}+6n-112=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -112 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-6±\sqrt{6^{2}-4\left(-112\right)}}{2}$$

Square 6.

$$n=\frac{-6±\sqrt{36-4\left(-112\right)}}{2}$$

Multiply -4 times -112.

$$n=\frac{-6±\sqrt{36+448}}{2}$$

Add 36 to 448.

$$n=\frac{-6±\sqrt{484}}{2}$$

Take the square root of 484.

$$n=\frac{-6±22}{2}$$

Now solve the equation n=\frac{-6±22}{2} when ± is plus. Add -6 to 22.

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

Divide 16 by 2.

$$n=8$$

Now solve the equation n=\frac{-6±22}{2} when ± is minus. Subtract 22 from -6.

$$n=-\frac{28}{2}$$

Divide -28 by 2.

$$n=-14$$

The equation is now solved.

$$n=8 n=-14$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(n+3\right)^{2}}=\sqrt{121}$$

Simplify.

$$n+3=11 n+3=-11$$

Subtract 3 from both sides of the equation.

$$n=8 n=-14$$