Solve (2x+5)^2-9=0

$$( 2 x + 5 ) ^ { 2 } - 9 = 0$$

Solve for x

$$x=-1$$

$$x=-4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+5\right)^{2}.

$$4x^{2}+20x+25-9=0$$

Subtract 9 from 25 to get 16.

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

Divide both sides by 4.

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

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

$$a+b=5 ab=1\times 4=4$$

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=1 b=4$$

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

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

Factor out x in the first and 4 in the second group.

$$x\left(x+1\right)+4\left(x+1\right)$$

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

$$\left(x+1\right)\left(x+4\right)$$

To find equation solutions, solve x+1=0 and x+4=0.

$$x=-1 x=-4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+5\right)^{2}.

$$4x^{2}+20x+25-9=0$$

Subtract 9 from 25 to get 16.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 20 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 20.

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

Multiply -4 times 4.

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

Multiply -16 times 16.

$$x=\frac{-20±\sqrt{400-256}}{2\times 4}$$

Add 400 to -256.

$$x=\frac{-20±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$x=\frac{-20±12}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-20±12}{8}$$

Now solve the equation x=\frac{-20±12}{8} when ± is plus. Add -20 to 12.

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

Divide -8 by 8.

$$x=-1$$

Now solve the equation x=\frac{-20±12}{8} when ± is minus. Subtract 12 from -20.

$$x=-\frac{32}{8}$$

Divide -32 by 8.

$$x=-4$$

The equation is now solved.

$$x=-1 x=-4$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+5\right)^{2}.

$$4x^{2}+20x+25-9=0$$

Subtract 9 from 25 to get 16.

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

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

$$\frac{4x^{2}+20x}{4}=-\frac{16}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{20}{4}x=-\frac{16}{4}$$

Divide 20 by 4.

$$x^{2}+5x=-\frac{16}{4}$$

Divide -16 by 4.

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

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.

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

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

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

Add -4 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{9}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

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

$$x=-1 x=-4$$