Solve (2x+4)^2=100

$$( 2 x + 4 ) ^ { 2 } = 100$$

Solve for x

$$x=3$$

$$x=-7$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+4\right)^{2}.

$$4x^{2}+16x+16=100$$

Subtract 100 from both sides.

$$4x^{2}+16x+16-100=0$$

Subtract 100 from 16 to get -84.

$$4x^{2}+16x-84=0$$

Divide both sides by 4.

$$x^{2}+4x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-21\right)=-21$$

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 -21.

$$-1,21 -3,7$$

Calculate the sum for each pair.

$$-1+21=20 -3+7=4$$

The solution is the pair that gives sum 4.

$$a=-3 b=7$$

Rewrite x^{2}+4x-21 as \left(x^{2}-3x\right)+\left(7x-21\right).

$$\left(x^{2}-3x\right)+\left(7x-21\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-3\right)+7\left(x-3\right)$$

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

$$\left(x-3\right)\left(x+7\right)$$

To find equation solutions, solve x-3=0 and x+7=0.

$$x=3 x=-7$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+4\right)^{2}.

$$4x^{2}+16x+16=100$$

Subtract 100 from both sides.

$$4x^{2}+16x+16-100=0$$

Subtract 100 from 16 to get -84.

$$4x^{2}+16x-84=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 16 for b, and -84 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 4\left(-84\right)}}{2\times 4}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 4\left(-84\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-16±\sqrt{256-16\left(-84\right)}}{2\times 4}$$

Multiply -16 times -84.

$$x=\frac{-16±\sqrt{256+1344}}{2\times 4}$$

Add 256 to 1344.

$$x=\frac{-16±\sqrt{1600}}{2\times 4}$$

Take the square root of 1600.

$$x=\frac{-16±40}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-16±40}{8}$$

Now solve the equation x=\frac{-16±40}{8} when ± is plus. Add -16 to 40.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{-16±40}{8} when ± is minus. Subtract 40 from -16.

$$x=-\frac{56}{8}$$

Divide -56 by 8.

$$x=-7$$

The equation is now solved.

$$x=3 x=-7$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+4\right)^{2}.

$$4x^{2}+16x+16=100$$

Subtract 16 from both sides.

$$4x^{2}+16x=100-16$$

Subtract 16 from 100 to get 84.

$$4x^{2}+16x=84$$

Divide both sides by 4.

$$\frac{4x^{2}+16x}{4}=\frac{84}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{16}{4}x=\frac{84}{4}$$

Divide 16 by 4.

$$x^{2}+4x=\frac{84}{4}$$

Divide 84 by 4.

$$x^{2}+4x=21$$

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.

$$x^{2}+4x+2^{2}=21+2^{2}$$

Square 2.

$$x^{2}+4x+4=21+4$$

Add 21 to 4.

$$x^{2}+4x+4=25$$

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}.

$$\left(x+2\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{25}$$

Simplify.

$$x+2=5 x+2=-5$$

Subtract 2 from both sides of the equation.

$$x=3 x=-7$$