Solve 4*left(x+2right)^2-1=0

$$4 \times { \left(x+2 \right) }^{ 2 } -1=0$$

Solve for x

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

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

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}.

$$4\left(x^{2}+4x+4\right)-1=0$$

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

$$4x^{2}+16x+16-1=0$$

Subtract 1 from 16 to get 15.

$$4x^{2}+16x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=16 ab=4\times 15=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 16.

$$a=6 b=10$$

Rewrite 4x^{2}+16x+15 as \left(4x^{2}+6x\right)+\left(10x+15\right).

$$\left(4x^{2}+6x\right)+\left(10x+15\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(2x+3\right)+5\left(2x+3\right)$$

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

$$\left(2x+3\right)\left(2x+5\right)$$

To find equation solutions, solve 2x+3=0 and 2x+5=0.

$$x=-\frac{3}{2} x=-\frac{5}{2}$$

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}.

$$4\left(x^{2}+4x+4\right)-1=0$$

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

$$4x^{2}+16x+16-1=0$$

Subtract 1 from 16 to get 15.

$$4x^{2}+16x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 16 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 4\times 15}}{2\times 4}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 4\times 15}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-16±\sqrt{256-16\times 15}}{2\times 4}$$

Multiply -16 times 15.

$$x=\frac{-16±\sqrt{256-240}}{2\times 4}$$

Add 256 to -240.

$$x=\frac{-16±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$x=\frac{-16±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-16±4}{8}$$

Now solve the equation x=\frac{-16±4}{8} when ± is plus. Add -16 to 4.

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

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{-16±4}{8} when ± is minus. Subtract 4 from -16.

$$x=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=-\frac{3}{2} x=-\frac{5}{2}$$

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}.

$$4\left(x^{2}+4x+4\right)-1=0$$

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

$$4x^{2}+16x+16-1=0$$

Subtract 1 from 16 to get 15.

$$4x^{2}+16x+15=0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+16x=-15$$

Divide both sides by 4.

$$\frac{4x^{2}+16x}{4}=-\frac{15}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{16}{4}x=-\frac{15}{4}$$

Divide 16 by 4.

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

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.

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

Square 2.

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

Add -\frac{15}{4} to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 2 from both sides of the equation.

$$x=-\frac{3}{2} x=-\frac{5}{2}$$