Solve (x+1)(2x+3)+(x+1)(3x-2)=0

$$( x + 1 ) ( 2 x + 3 ) + ( x + 1 ) ( 3 x - 2 ) = 0$$

Solve for x

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

$$x=-1$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x+1 by 2x+3 and combine like terms.

$$2x^{2}+5x+3+\left(x+1\right)\left(3x-2\right)=0$$

Use the distributive property to multiply x+1 by 3x-2 and combine like terms.

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

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

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

Combine 5x and x to get 6x.

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

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 the distributive property to multiply x+1 by 2x+3 and combine like terms.

$$2x^{2}+5x+3+\left(x+1\right)\left(3x-2\right)=0$$

Use the distributive property to multiply x+1 by 3x-2 and combine like terms.

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

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

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

Combine 5x and x to get 6x.

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

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 the distributive property to multiply x+1 by 2x+3 and combine like terms.

$$2x^{2}+5x+3+\left(x+1\right)\left(3x-2\right)=0$$

Use the distributive property to multiply x+1 by 3x-2 and combine like terms.

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

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

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

Combine 5x and x to get 6x.

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

Subtract 2 from 3 to get 1.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

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.

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