Solve (2x+5)^2=(x+2)^2

$$( 2 x + 5 ) ^ { 2 } = ( x + 2 ) ^ { 2 }$$

Solve for x

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$x=-3$$

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}.

$$4x^{2}+20x+25=\left(x+2\right)^{2}$$

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

$$4x^{2}+20x+25=x^{2}+4x+4$$

Subtract x^{2} from both sides.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+20x+25=4x+4$$

Subtract 4x from both sides.

$$3x^{2}+20x+25-4x=4$$

Combine 20x and -4x to get 16x.

$$3x^{2}+16x+25=4$$

Subtract 4 from both sides.

$$3x^{2}+16x+25-4=0$$

Subtract 4 from 25 to get 21.

$$3x^{2}+16x+21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=16 ab=3\times 21=63$$

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 63.

$$1,63 3,21 7,9$$

Calculate the sum for each pair.

$$1+63=64 3+21=24 7+9=16$$

The solution is the pair that gives sum 16.

$$a=7 b=9$$

Rewrite 3x^{2}+16x+21 as \left(3x^{2}+7x\right)+\left(9x+21\right).

$$\left(3x^{2}+7x\right)+\left(9x+21\right)$$

Factor out x in the first and 3 in the second group.

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

Factor out common term 3x+7 by using distributive property.

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

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

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

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}.

$$4x^{2}+20x+25=\left(x+2\right)^{2}$$

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

$$4x^{2}+20x+25=x^{2}+4x+4$$

Subtract x^{2} from both sides.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+20x+25=4x+4$$

Subtract 4x from both sides.

$$3x^{2}+20x+25-4x=4$$

Combine 20x and -4x to get 16x.

$$3x^{2}+16x+25=4$$

Subtract 4 from both sides.

$$3x^{2}+16x+25-4=0$$

Subtract 4 from 25 to get 21.

$$3x^{2}+16x+21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 16 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 3\times 21}}{2\times 3}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 3\times 21}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-16±\sqrt{256-12\times 21}}{2\times 3}$$

Multiply -12 times 21.

$$x=\frac{-16±\sqrt{256-252}}{2\times 3}$$

Add 256 to -252.

$$x=\frac{-16±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$x=\frac{-16±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-16±2}{6}$$

Now solve the equation x=\frac{-16±2}{6} when ± is plus. Add -16 to 2.

$$x=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-16±2}{6} when ± is minus. Subtract 2 from -16.

$$x=-\frac{18}{6}$$

Divide -18 by 6.

$$x=-3$$

The equation is now solved.

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

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}.

$$4x^{2}+20x+25=\left(x+2\right)^{2}$$

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

$$4x^{2}+20x+25=x^{2}+4x+4$$

Subtract x^{2} from both sides.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+20x+25=4x+4$$

Subtract 4x from both sides.

$$3x^{2}+20x+25-4x=4$$

Combine 20x and -4x to get 16x.

$$3x^{2}+16x+25=4$$

Subtract 25 from both sides.

$$3x^{2}+16x=4-25$$

Subtract 25 from 4 to get -21.

$$3x^{2}+16x=-21$$

Divide both sides by 3.

$$\frac{3x^{2}+16x}{3}=-\frac{21}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{16}{3}x=-\frac{21}{3}$$

Divide -21 by 3.

$$x^{2}+\frac{16}{3}x=-7$$

Divide \frac{16}{3}, the coefficient of the x term, by 2 to get \frac{8}{3}. Then add the square of \frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{16}{3}x+\left(\frac{8}{3}\right)^{2}=-7+\left(\frac{8}{3}\right)^{2}$$

Square \frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{16}{3}x+\frac{64}{9}=-7+\frac{64}{9}$$

Add -7 to \frac{64}{9}.

$$x^{2}+\frac{16}{3}x+\frac{64}{9}=\frac{1}{9}$$

Factor x^{2}+\frac{16}{3}x+\frac{64}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{8}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{8}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x+\frac{8}{3}=\frac{1}{3} x+\frac{8}{3}=-\frac{1}{3}$$

Subtract \frac{8}{3} from both sides of the equation.

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