Solve (x+5)^2-36=0

$$( x + 5 ) ^ { 2 } - 36 = 0$$

Solve for x

$$x=1$$

$$x=-11$$

Steps Using Factoring

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

$$x^{2}+10x+25-36=0$$

Subtract 36 from 25 to get -11.

$$x^{2}+10x-11=0$$

To solve the equation, factor x^{2}+10x-11 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.

$$a+b=10 ab=-11$$

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.

$$a=-1 b=11$$

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

$$\left(x-1\right)\left(x+11\right)$$

To find equation solutions, solve x-1=0 and x+11=0.

$$x=1 x=-11$$

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}.

$$x^{2}+10x+25-36=0$$

Subtract 36 from 25 to get -11.

$$x^{2}+10x-11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-11. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-11\right)=-11$$

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.

$$a=-1 b=11$$

Rewrite x^{2}+10x-11 as \left(x^{2}-x\right)+\left(11x-11\right).

$$\left(x^{2}-x\right)+\left(11x-11\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(x-1\right)+11\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+11\right)$$

To find equation solutions, solve x-1=0 and x+11=0.

$$x=1 x=-11$$

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}.

$$x^{2}+10x+25-36=0$$

Subtract 36 from 25 to get -11.

$$x^{2}+10x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-11\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$x=\frac{-10±\sqrt{100+44}}{2}$$

Add 100 to 44.

$$x=\frac{-10±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-10±12}{2}$$

Now solve the equation x=\frac{-10±12}{2} when ± is plus. Add -10 to 12.

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

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-10±12}{2} when ± is minus. Subtract 12 from -10.

$$x=-\frac{22}{2}$$

Divide -22 by 2.

$$x=-11$$

The equation is now solved.

$$x=1 x=-11$$

Steps for Completing the Square

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

$$x^{2}+10x+25-36=0$$

Subtract 36 from 25 to get -11.

$$x^{2}+10x-11=0$$

Add 11 to both sides. Anything plus zero gives itself.

$$x^{2}+10x=11$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=11+5^{2}$$

Square 5.

$$x^{2}+10x+25=11+25$$

Add 11 to 25.

$$x^{2}+10x+25=36$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+5=6 x+5=-6$$

Subtract 5 from both sides of the equation.

$$x=1 x=-11$$