Solve 3(x+2)^2-27=0

$$3 ( x + 2 ) ^ { 2 } - 27 = 0$$

Solve for x

$$x=1$$

$$x=-5$$

Steps Using Factoring By Grouping

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

$$3\left(x^{2}+4x+4\right)-27=0$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-27=0$$

Subtract 27 from 12 to get -15.

$$3x^{2}+12x-15=0$$

Divide both sides by 3.

$$x^{2}+4x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-5\right)=-5$$

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.

$$a=-1 b=5$$

Rewrite x^{2}+4x-5 as \left(x^{2}-x\right)+\left(5x-5\right).

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

Factor out x in the first and 5 in the second group.

$$x\left(x-1\right)+5\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve x-1=0 and x+5=0.

$$x=1 x=-5$$

Steps Using the Quadratic Formula

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

$$3\left(x^{2}+4x+4\right)-27=0$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-27=0$$

Subtract 27 from 12 to get -15.

$$3x^{2}+12x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 12 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 3\left(-15\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-12±\sqrt{144-12\left(-15\right)}}{2\times 3}$$

Multiply -12 times -15.

$$x=\frac{-12±\sqrt{144+180}}{2\times 3}$$

Add 144 to 180.

$$x=\frac{-12±\sqrt{324}}{2\times 3}$$

Take the square root of 324.

$$x=\frac{-12±18}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-12±18}{6}$$

Now solve the equation x=\frac{-12±18}{6} when ± is plus. Add -12 to 18.

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

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{-12±18}{6} when ± is minus. Subtract 18 from -12.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

The equation is now solved.

$$x=1 x=-5$$

Steps for Completing the Square

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

$$3\left(x^{2}+4x+4\right)-27=0$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

$$3x^{2}+12x+12-27=0$$

Subtract 27 from 12 to get -15.

$$3x^{2}+12x-15=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$3x^{2}+12x=15$$

Divide both sides by 3.

$$\frac{3x^{2}+12x}{3}=\frac{15}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{12}{3}x=\frac{15}{3}$$

Divide 12 by 3.

$$x^{2}+4x=\frac{15}{3}$$

Divide 15 by 3.

$$x^{2}+4x=5$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=5+2^{2}$$

Square 2.

$$x^{2}+4x+4=5+4$$

Add 5 to 4.

$$x^{2}+4x+4=9$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+2=3 x+2=-3$$

Subtract 2 from both sides of the equation.

$$x=1 x=-5$$