Solve (2x+2)^2=81

$$( 2 x + 2 ) ^ { 2 } = 81$$

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$x = -\frac{11}{2} = -5\frac{1}{2} = -5.5$$

Steps Using Factoring By Grouping

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

$$4x^{2}+8x+4=81$$

Subtract 81 from both sides.

$$4x^{2}+8x+4-81=0$$

Subtract 81 from 4 to get -77.

$$4x^{2}+8x-77=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-77. To find a and b, set up a system to be solved.

$$a+b=8 ab=4\left(-77\right)=-308$$

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 -308.

$$-1,308 -2,154 -4,77 -7,44 -11,28 -14,22$$

Calculate the sum for each pair.

$$-1+308=307 -2+154=152 -4+77=73 -7+44=37 -11+28=17 -14+22=8$$

The solution is the pair that gives sum 8.

$$a=-14 b=22$$

Rewrite 4x^{2}+8x-77 as \left(4x^{2}-14x\right)+\left(22x-77\right).

$$\left(4x^{2}-14x\right)+\left(22x-77\right)$$

Factor out 2x in the first and 11 in the second group.

$$2x\left(2x-7\right)+11\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(2x+11\right)$$

To find equation solutions, solve 2x-7=0 and 2x+11=0.

$$x=\frac{7}{2} x=-\frac{11}{2}$$

Steps Using the Quadratic Formula

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

$$4x^{2}+8x+4=81$$

Subtract 81 from both sides.

$$4x^{2}+8x+4-81=0$$

Subtract 81 from 4 to get -77.

$$4x^{2}+8x-77=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 8 for b, and -77 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\left(-77\right)}}{2\times 4}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 4\left(-77\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-8±\sqrt{64-16\left(-77\right)}}{2\times 4}$$

Multiply -16 times -77.

$$x=\frac{-8±\sqrt{64+1232}}{2\times 4}$$

Add 64 to 1232.

$$x=\frac{-8±\sqrt{1296}}{2\times 4}$$

Take the square root of 1296.

$$x=\frac{-8±36}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-8±36}{8}$$

Now solve the equation x=\frac{-8±36}{8} when ± is plus. Add -8 to 36.

$$x=\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{-8±36}{8} when ± is minus. Subtract 36 from -8.

$$x=-\frac{44}{8}$$

Reduce the fraction \frac{-44}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{11}{2}$$

The equation is now solved.

$$x=\frac{7}{2} x=-\frac{11}{2}$$

Steps for Completing the Square

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

$$4x^{2}+8x+4=81$$

Subtract 4 from both sides.

$$4x^{2}+8x=81-4$$

Subtract 4 from 81 to get 77.

$$4x^{2}+8x=77$$

Divide both sides by 4.

$$\frac{4x^{2}+8x}{4}=\frac{77}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{8}{4}x=\frac{77}{4}$$

Divide 8 by 4.

$$x^{2}+2x=\frac{77}{4}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{77}{4}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{77}{4}+1$$

Add \frac{77}{4} to 1.

$$x^{2}+2x+1=\frac{81}{4}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x+1=\frac{9}{2} x+1=-\frac{9}{2}$$

Subtract 1 from both sides of the equation.

$$x=\frac{7}{2} x=-\frac{11}{2}$$