Solve (x+3)^2=36

$$( x + 3 ) ^ { 2 } = 36$$

Solve for x

$$x=3$$

$$x=-9$$

Steps Using Factoring

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

$$x^{2}+6x+9=36$$

Subtract 36 from both sides.

$$x^{2}+6x+9-36=0$$

Subtract 36 from 9 to get -27.

$$x^{2}+6x-27=0$$

To solve the equation, factor x^{2}+6x-27 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.

$$a+b=6 ab=-27$$

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 -27.

$$-1,27 -3,9$$

Calculate the sum for each pair.

$$-1+27=26 -3+9=6$$

The solution is the pair that gives sum 6.

$$a=-3 b=9$$

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

$$\left(x-3\right)\left(x+9\right)$$

To find equation solutions, solve x-3=0 and x+9=0.

$$x=3 x=-9$$

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}.

$$x^{2}+6x+9=36$$

Subtract 36 from both sides.

$$x^{2}+6x+9-36=0$$

Subtract 36 from 9 to get -27.

$$x^{2}+6x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-27\right)=-27$$

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 -27.

$$-1,27 -3,9$$

Calculate the sum for each pair.

$$-1+27=26 -3+9=6$$

The solution is the pair that gives sum 6.

$$a=-3 b=9$$

Rewrite x^{2}+6x-27 as \left(x^{2}-3x\right)+\left(9x-27\right).

$$\left(x^{2}-3x\right)+\left(9x-27\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x-3\right)+9\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+9\right)$$

To find equation solutions, solve x-3=0 and x+9=0.

$$x=3 x=-9$$

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}.

$$x^{2}+6x+9=36$$

Subtract 36 from both sides.

$$x^{2}+6x+9-36=0$$

Subtract 36 from 9 to get -27.

$$x^{2}+6x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-27\right)}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-27\right)}}{2}$$

Multiply -4 times -27.

$$x=\frac{-6±\sqrt{36+108}}{2}$$

Add 36 to 108.

$$x=\frac{-6±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-6±12}{2}$$

Now solve the equation x=\frac{-6±12}{2} when ± is plus. Add -6 to 12.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-6±12}{2} when ± is minus. Subtract 12 from -6.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=3 x=-9$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+3=6 x+3=-6$$

Subtract 3 from both sides of the equation.

$$x=3 x=-9$$