Solve (2x+4)^2+2(2x+4)-3=0

$$( 2 x + 4 ) ^ { 2 } + 2 ( 2 x + 4 ) - 3 = 0$$

Solve for x

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

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

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}.

$$4x^{2}+16x+16+2\left(2x+4\right)-3=0$$

Use the distributive property to multiply 2 by 2x+4.

$$4x^{2}+16x+16+4x+8-3=0$$

Combine 16x and 4x to get 20x.

$$4x^{2}+20x+16+8-3=0$$

Add 16 and 8 to get 24.

$$4x^{2}+20x+24-3=0$$

Subtract 3 from 24 to get 21.

$$4x^{2}+20x+21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=20 ab=4\times 21=84$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 84.

$$1,84 2,42 3,28 4,21 6,14 7,12$$

Calculate the sum for each pair.

$$1+84=85 2+42=44 3+28=31 4+21=25 6+14=20 7+12=19$$

The solution is the pair that gives sum 20.

$$a=6 b=14$$

Rewrite 4x^{2}+20x+21 as \left(4x^{2}+6x\right)+\left(14x+21\right).

$$\left(4x^{2}+6x\right)+\left(14x+21\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(2x+3\right)+7\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(2x+7\right)$$

To find equation solutions, solve 2x+3=0 and 2x+7=0.

$$x=-\frac{3}{2} x=-\frac{7}{2}$$

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}.

$$4x^{2}+16x+16+2\left(2x+4\right)-3=0$$

Use the distributive property to multiply 2 by 2x+4.

$$4x^{2}+16x+16+4x+8-3=0$$

Combine 16x and 4x to get 20x.

$$4x^{2}+20x+16+8-3=0$$

Add 16 and 8 to get 24.

$$4x^{2}+20x+24-3=0$$

Subtract 3 from 24 to get 21.

$$4x^{2}+20x+21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 20 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 4\times 21}}{2\times 4}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 4\times 21}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-20±\sqrt{400-16\times 21}}{2\times 4}$$

Multiply -16 times 21.

$$x=\frac{-20±\sqrt{400-336}}{2\times 4}$$

Add 400 to -336.

$$x=\frac{-20±\sqrt{64}}{2\times 4}$$

Take the square root of 64.

$$x=\frac{-20±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-20±8}{8}$$

Now solve the equation x=\frac{-20±8}{8} when ± is plus. Add -20 to 8.

$$x=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{3}{2}$$

Now solve the equation x=\frac{-20±8}{8} when ± is minus. Subtract 8 from -20.

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

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

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

The equation is now solved.

$$x=-\frac{3}{2} x=-\frac{7}{2}$$

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}.

$$4x^{2}+16x+16+2\left(2x+4\right)-3=0$$

Use the distributive property to multiply 2 by 2x+4.

$$4x^{2}+16x+16+4x+8-3=0$$

Combine 16x and 4x to get 20x.

$$4x^{2}+20x+16+8-3=0$$

Add 16 and 8 to get 24.

$$4x^{2}+20x+24-3=0$$

Subtract 3 from 24 to get 21.

$$4x^{2}+20x+21=0$$

Subtract 21 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+20x=-21$$

Divide both sides by 4.

$$\frac{4x^{2}+20x}{4}=-\frac{21}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{20}{4}x=-\frac{21}{4}$$

Divide 20 by 4.

$$x^{2}+5x=-\frac{21}{4}$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{21}{4}+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=\frac{-21+25}{4}$$

Add -\frac{21}{4} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+5x+\frac{25}{4}=1$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+\frac{5}{2}=1 x+\frac{5}{2}=-1$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=-\frac{3}{2} x=-\frac{7}{2}$$