Solve (2x+1)^2+(x+2)(x+1)=x+2

$$( 2 x + 1 ) ^ { 2 } + ( x + 2 ) ( x + 1 ) = x + 2$$

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

Use the distributive property to multiply x+2 by x+1 and combine like terms.

$$4x^{2}+4x+1+x^{2}+3x+2=x+2$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}+4x+1+3x+2=x+2$$

Combine 4x and 3x to get 7x.

$$5x^{2}+7x+1+2=x+2$$

Add 1 and 2 to get 3.

$$5x^{2}+7x+3=x+2$$

Subtract x from both sides.

$$5x^{2}+7x+3-x=2$$

Combine 7x and -x to get 6x.

$$5x^{2}+6x+3=2$$

Subtract 2 from both sides.

$$5x^{2}+6x+3-2=0$$

Subtract 2 from 3 to get 1.

$$5x^{2}+6x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=6 ab=5\times 1=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=5$$

Rewrite 5x^{2}+6x+1 as \left(5x^{2}+x\right)+\left(5x+1\right).

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

Factor out x in 5x^{2}+x.

$$x\left(5x+1\right)+5x+1$$

Factor out common term 5x+1 by using distributive property.

$$\left(5x+1\right)\left(x+1\right)$$

To find equation solutions, solve 5x+1=0 and x+1=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

Use the distributive property to multiply x+2 by x+1 and combine like terms.

$$4x^{2}+4x+1+x^{2}+3x+2=x+2$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}+4x+1+3x+2=x+2$$

Combine 4x and 3x to get 7x.

$$5x^{2}+7x+1+2=x+2$$

Add 1 and 2 to get 3.

$$5x^{2}+7x+3=x+2$$

Subtract x from both sides.

$$5x^{2}+7x+3-x=2$$

Combine 7x and -x to get 6x.

$$5x^{2}+6x+3=2$$

Subtract 2 from both sides.

$$5x^{2}+6x+3-2=0$$

Subtract 2 from 3 to get 1.

$$5x^{2}+6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5}}{2\times 5}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-6±\sqrt{36-20}}{2\times 5}$$

Add 36 to -20.

$$x=\frac{-6±\sqrt{16}}{2\times 5}$$

Take the square root of 16.

$$x=\frac{-6±4}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-6±4}{10}$$

Now solve the equation x=\frac{-6±4}{10} when ± is plus. Add -6 to 4.

$$x=-\frac{2}{10}$$

Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-6±4}{10} when ± is minus. Subtract 4 from -6.

$$x=-\frac{10}{10}$$

Divide -10 by 10.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

Use the distributive property to multiply x+2 by x+1 and combine like terms.

$$4x^{2}+4x+1+x^{2}+3x+2=x+2$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}+4x+1+3x+2=x+2$$

Combine 4x and 3x to get 7x.

$$5x^{2}+7x+1+2=x+2$$

Add 1 and 2 to get 3.

$$5x^{2}+7x+3=x+2$$

Subtract x from both sides.

$$5x^{2}+7x+3-x=2$$

Combine 7x and -x to get 6x.

$$5x^{2}+6x+3=2$$

Subtract 3 from both sides.

$$5x^{2}+6x=2-3$$

Subtract 3 from 2 to get -1.

$$5x^{2}+6x=-1$$

Divide both sides by 5.

$$\frac{5x^{2}+6x}{5}=-\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{6}{5}x=-\frac{1}{5}$$

Divide \frac{6}{5}, the coefficient of the x term, by 2 to get \frac{3}{5}. Then add the square of \frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=-\frac{1}{5}+\left(\frac{3}{5}\right)^{2}$$

Square \frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=-\frac{1}{5}+\frac{9}{25}$$

Add -\frac{1}{5} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{4}{25}$$

Factor x^{2}+\frac{6}{5}x+\frac{9}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

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

Subtract \frac{3}{5} from both sides of the equation.

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