Solve (x+7)^2=36

$$( x + 7 ) ^ { 2 } = 36$$

Solve for x

$$x=-1$$

$$x=-13$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$x^{2}+14x+49=36$$

Subtract 36 from both sides.

$$x^{2}+14x+49-36=0$$

Subtract 36 from 49 to get 13.

$$x^{2}+14x+13=0$$

To solve the equation, factor x^{2}+14x+13 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.

$$a+b=14 ab=13$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=13$$

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

$$\left(x+1\right)\left(x+13\right)$$

To find equation solutions, solve x+1=0 and x+13=0.

$$x=-1 x=-13$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$x^{2}+14x+49=36$$

Subtract 36 from both sides.

$$x^{2}+14x+49-36=0$$

Subtract 36 from 49 to get 13.

$$x^{2}+14x+13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+13. To find a and b, set up a system to be solved.

$$a+b=14 ab=1\times 13=13$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=13$$

Rewrite x^{2}+14x+13 as \left(x^{2}+x\right)+\left(13x+13\right).

$$\left(x^{2}+x\right)+\left(13x+13\right)$$

Factor out x in the first and 13 in the second group.

$$x\left(x+1\right)+13\left(x+1\right)$$

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

$$\left(x+1\right)\left(x+13\right)$$

To find equation solutions, solve x+1=0 and x+13=0.

$$x=-1 x=-13$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$x^{2}+14x+49=36$$

Subtract 36 from both sides.

$$x^{2}+14x+49-36=0$$

Subtract 36 from 49 to get 13.

$$x^{2}+14x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 14 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\times 13}}{2}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 13}}{2}$$

Multiply -4 times 13.

$$x=\frac{-14±\sqrt{196-52}}{2}$$

Add 196 to -52.

$$x=\frac{-14±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-14±12}{2}$$

Now solve the equation x=\frac{-14±12}{2} when ± is plus. Add -14 to 12.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-14±12}{2} when ± is minus. Subtract 12 from -14.

$$x=-\frac{26}{2}$$

Divide -26 by 2.

$$x=-13$$

The equation is now solved.

$$x=-1 x=-13$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x+7\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+7=6 x+7=-6$$

Subtract 7 from both sides of the equation.

$$x=-1 x=-13$$