Solve (x+5)^2=121

$$( x + 5 ) ^ { 2 } = 121$$

Solve for x

$$x=6$$

$$x=-16$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=121$$

Subtract 121 from both sides.

$$x^{2}+10x+25-121=0$$

Subtract 121 from 25 to get -96.

$$x^{2}+10x-96=0$$

To solve the equation, factor x^{2}+10x-96 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.

$$a+b=10 ab=-96$$

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 -96.

$$-1,96 -2,48 -3,32 -4,24 -6,16 -8,12$$

Calculate the sum for each pair.

$$-1+96=95 -2+48=46 -3+32=29 -4+24=20 -6+16=10 -8+12=4$$

The solution is the pair that gives sum 10.

$$a=-6 b=16$$

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

$$\left(x-6\right)\left(x+16\right)$$

To find equation solutions, solve x-6=0 and x+16=0.

$$x=6 x=-16$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=121$$

Subtract 121 from both sides.

$$x^{2}+10x+25-121=0$$

Subtract 121 from 25 to get -96.

$$x^{2}+10x-96=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-96. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-96\right)=-96$$

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 -96.

$$-1,96 -2,48 -3,32 -4,24 -6,16 -8,12$$

Calculate the sum for each pair.

$$-1+96=95 -2+48=46 -3+32=29 -4+24=20 -6+16=10 -8+12=4$$

The solution is the pair that gives sum 10.

$$a=-6 b=16$$

Rewrite x^{2}+10x-96 as \left(x^{2}-6x\right)+\left(16x-96\right).

$$\left(x^{2}-6x\right)+\left(16x-96\right)$$

Factor out x in the first and 16 in the second group.

$$x\left(x-6\right)+16\left(x-6\right)$$

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

$$\left(x-6\right)\left(x+16\right)$$

To find equation solutions, solve x-6=0 and x+16=0.

$$x=6 x=-16$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=121$$

Subtract 121 from both sides.

$$x^{2}+10x+25-121=0$$

Subtract 121 from 25 to get -96.

$$x^{2}+10x-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-96\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-96\right)}}{2}$$

Multiply -4 times -96.

$$x=\frac{-10±\sqrt{100+384}}{2}$$

Add 100 to 384.

$$x=\frac{-10±\sqrt{484}}{2}$$

Take the square root of 484.

$$x=\frac{-10±22}{2}$$

Now solve the equation x=\frac{-10±22}{2} when ± is plus. Add -10 to 22.

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

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{-10±22}{2} when ± is minus. Subtract 22 from -10.

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

Divide -32 by 2.

$$x=-16$$

The equation is now solved.

$$x=6 x=-16$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{121}$$

Simplify.

$$x+5=11 x+5=-11$$

Subtract 5 from both sides of the equation.

$$x=6 x=-16$$