Solve (2+n)(7n-1)+(2+n)(3n+4)=0

$$( 2 + n ) ( 7 n - 1 ) + ( 2 + n ) ( 3 n + 4 ) = 0$$

Solve for n

$$n=-\frac{3}{10}=-0.3$$

$$n=-2$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2+n by 7n-1 and combine like terms.

$$13n-2+7n^{2}+\left(2+n\right)\left(3n+4\right)=0$$

Use the distributive property to multiply 2+n by 3n+4 and combine like terms.

$$13n-2+7n^{2}+10n+8+3n^{2}=0$$

Combine 13n and 10n to get 23n.

$$23n-2+7n^{2}+8+3n^{2}=0$$

Add -2 and 8 to get 6.

$$23n+6+7n^{2}+3n^{2}=0$$

Combine 7n^{2} and 3n^{2} to get 10n^{2}.

$$23n+6+10n^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$10n^{2}+23n+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10n^{2}+an+bn+6. To find a and b, set up a system to be solved.

$$a+b=23 ab=10\times 6=60$$

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 23.

$$a=3 b=20$$

Rewrite 10n^{2}+23n+6 as \left(10n^{2}+3n\right)+\left(20n+6\right).

$$\left(10n^{2}+3n\right)+\left(20n+6\right)$$

Factor out n in the first and 2 in the second group.

$$n\left(10n+3\right)+2\left(10n+3\right)$$

Factor out common term 10n+3 by using distributive property.

$$\left(10n+3\right)\left(n+2\right)$$

To find equation solutions, solve 10n+3=0 and n+2=0.

$$n=-\frac{3}{10} n=-2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2+n by 7n-1 and combine like terms.

$$13n-2+7n^{2}+\left(2+n\right)\left(3n+4\right)=0$$

Use the distributive property to multiply 2+n by 3n+4 and combine like terms.

$$13n-2+7n^{2}+10n+8+3n^{2}=0$$

Combine 13n and 10n to get 23n.

$$23n-2+7n^{2}+8+3n^{2}=0$$

Add -2 and 8 to get 6.

$$23n+6+7n^{2}+3n^{2}=0$$

Combine 7n^{2} and 3n^{2} to get 10n^{2}.

$$23n+6+10n^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$10n^{2}+23n+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 23 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-23±\sqrt{23^{2}-4\times 10\times 6}}{2\times 10}$$

Square 23.

$$n=\frac{-23±\sqrt{529-4\times 10\times 6}}{2\times 10}$$

Multiply -4 times 10.

$$n=\frac{-23±\sqrt{529-40\times 6}}{2\times 10}$$

Multiply -40 times 6.

$$n=\frac{-23±\sqrt{529-240}}{2\times 10}$$

Add 529 to -240.

$$n=\frac{-23±\sqrt{289}}{2\times 10}$$

Take the square root of 289.

$$n=\frac{-23±17}{2\times 10}$$

Multiply 2 times 10.

$$n=\frac{-23±17}{20}$$

Now solve the equation n=\frac{-23±17}{20} when ± is plus. Add -23 to 17.

$$n=-\frac{6}{20}$$

Reduce the fraction \frac{-6}{20} to lowest terms by extracting and canceling out 2.

$$n=-\frac{3}{10}$$

Now solve the equation n=\frac{-23±17}{20} when ± is minus. Subtract 17 from -23.

$$n=-\frac{40}{20}$$

Divide -40 by 20.

$$n=-2$$

The equation is now solved.

$$n=-\frac{3}{10} n=-2$$

Steps for Completing the Square

Use the distributive property to multiply 2+n by 7n-1 and combine like terms.

$$13n-2+7n^{2}+\left(2+n\right)\left(3n+4\right)=0$$

Use the distributive property to multiply 2+n by 3n+4 and combine like terms.

$$13n-2+7n^{2}+10n+8+3n^{2}=0$$

Combine 13n and 10n to get 23n.

$$23n-2+7n^{2}+8+3n^{2}=0$$

Add -2 and 8 to get 6.

$$23n+6+7n^{2}+3n^{2}=0$$

Combine 7n^{2} and 3n^{2} to get 10n^{2}.

$$23n+6+10n^{2}=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$23n+10n^{2}=-6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$10n^{2}+23n=-6$$

Divide both sides by 10.

$$\frac{10n^{2}+23n}{10}=-\frac{6}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$n^{2}+\frac{23}{10}n=-\frac{6}{10}$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

$$n^{2}+\frac{23}{10}n=-\frac{3}{5}$$

Divide \frac{23}{10}, the coefficient of the x term, by 2 to get \frac{23}{20}. Then add the square of \frac{23}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{23}{10}n+\left(\frac{23}{20}\right)^{2}=-\frac{3}{5}+\left(\frac{23}{20}\right)^{2}$$

Square \frac{23}{20} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{23}{10}n+\frac{529}{400}=-\frac{3}{5}+\frac{529}{400}$$

Add -\frac{3}{5} to \frac{529}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{23}{10}n+\frac{529}{400}=\frac{289}{400}$$

Factor n^{2}+\frac{23}{10}n+\frac{529}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{23}{20}\right)^{2}=\frac{289}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{23}{20}\right)^{2}}=\sqrt{\frac{289}{400}}$$

Simplify.

$$n+\frac{23}{20}=\frac{17}{20} n+\frac{23}{20}=-\frac{17}{20}$$

Subtract \frac{23}{20} from both sides of the equation.

$$n=-\frac{3}{10} n=-2$$