Solve (3/c+2)(c+2)=((-5c/c+2)(c+2)

$$( \frac { 3 } { c + 2 } ) ( c + 2 ) = ( ( - 5 \frac { c } { c + 2 } ) ( c + 2 )$$

Solve for c

$$c=-\frac{3}{5}=-0.6$$

Steps Using Factoring By Grouping

Variable c cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by c+2.

$$3\left(c+2\right)=-5c\left(c+2\right)$$

Use the distributive property to multiply 3 by c+2.

$$3c+6=-5c\left(c+2\right)$$

Use the distributive property to multiply -5c by c+2.

$$3c+6=-5c^{2}-10c$$

Add 5c^{2} to both sides.

$$3c+6+5c^{2}=-10c$$

Add 10c to both sides.

$$3c+6+5c^{2}+10c=0$$

Combine 3c and 10c to get 13c.

$$13c+6+5c^{2}=0$$

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

$$5c^{2}+13c+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5c^{2}+ac+bc+6. To find a and b, set up a system to be solved.

$$a+b=13 ab=5\times 6=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 13.

$$a=3 b=10$$

Rewrite 5c^{2}+13c+6 as \left(5c^{2}+3c\right)+\left(10c+6\right).

$$\left(5c^{2}+3c\right)+\left(10c+6\right)$$

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

$$c\left(5c+3\right)+2\left(5c+3\right)$$

Factor out common term 5c+3 by using distributive property.

$$\left(5c+3\right)\left(c+2\right)$$

To find equation solutions, solve 5c+3=0 and c+2=0.

$$c=-\frac{3}{5} c=-2$$

Variable c cannot be equal to -2.

$$c=-\frac{3}{5}$$

Steps Using the Quadratic Formula

Variable c cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by c+2.

$$3\left(c+2\right)=-5c\left(c+2\right)$$

Use the distributive property to multiply 3 by c+2.

$$3c+6=-5c\left(c+2\right)$$

Use the distributive property to multiply -5c by c+2.

$$3c+6=-5c^{2}-10c$$

Add 5c^{2} to both sides.

$$3c+6+5c^{2}=-10c$$

Add 10c to both sides.

$$3c+6+5c^{2}+10c=0$$

Combine 3c and 10c to get 13c.

$$13c+6+5c^{2}=0$$

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.

$$5c^{2}+13c+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 13 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-13±\sqrt{13^{2}-4\times 5\times 6}}{2\times 5}$$

Square 13.

$$c=\frac{-13±\sqrt{169-4\times 5\times 6}}{2\times 5}$$

Multiply -4 times 5.

$$c=\frac{-13±\sqrt{169-20\times 6}}{2\times 5}$$

Multiply -20 times 6.

$$c=\frac{-13±\sqrt{169-120}}{2\times 5}$$

Add 169 to -120.

$$c=\frac{-13±\sqrt{49}}{2\times 5}$$

Take the square root of 49.

$$c=\frac{-13±7}{2\times 5}$$

Multiply 2 times 5.

$$c=\frac{-13±7}{10}$$

Now solve the equation c=\frac{-13±7}{10} when ± is plus. Add -13 to 7.

$$c=-\frac{6}{10}$$

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

$$c=-\frac{3}{5}$$

Now solve the equation c=\frac{-13±7}{10} when ± is minus. Subtract 7 from -13.

$$c=-\frac{20}{10}$$

Divide -20 by 10.

$$c=-2$$

The equation is now solved.

$$c=-\frac{3}{5} c=-2$$

Variable c cannot be equal to -2.

$$c=-\frac{3}{5}$$

Steps for Completing the Square

Variable c cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by c+2.

$$3\left(c+2\right)=-5c\left(c+2\right)$$

Use the distributive property to multiply 3 by c+2.

$$3c+6=-5c\left(c+2\right)$$

Use the distributive property to multiply -5c by c+2.

$$3c+6=-5c^{2}-10c$$

Add 5c^{2} to both sides.

$$3c+6+5c^{2}=-10c$$

Add 10c to both sides.

$$3c+6+5c^{2}+10c=0$$

Combine 3c and 10c to get 13c.

$$13c+6+5c^{2}=0$$

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

$$13c+5c^{2}=-6$$

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.

$$5c^{2}+13c=-6$$

Divide both sides by 5.

$$\frac{5c^{2}+13c}{5}=-\frac{6}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$c^{2}+\frac{13}{5}c=-\frac{6}{5}$$

Divide \frac{13}{5}, the coefficient of the x term, by 2 to get \frac{13}{10}. Then add the square of \frac{13}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}+\frac{13}{5}c+\left(\frac{13}{10}\right)^{2}=-\frac{6}{5}+\left(\frac{13}{10}\right)^{2}$$

Square \frac{13}{10} by squaring both the numerator and the denominator of the fraction.

$$c^{2}+\frac{13}{5}c+\frac{169}{100}=-\frac{6}{5}+\frac{169}{100}$$

Add -\frac{6}{5} to \frac{169}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$c^{2}+\frac{13}{5}c+\frac{169}{100}=\frac{49}{100}$$

Factor c^{2}+\frac{13}{5}c+\frac{169}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c+\frac{13}{10}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c+\frac{13}{10}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$c+\frac{13}{10}=\frac{7}{10} c+\frac{13}{10}=-\frac{7}{10}$$

Subtract \frac{13}{10} from both sides of the equation.

$$c=-\frac{3}{5} c=-2$$

Variable c cannot be equal to -2.

$$c=-\frac{3}{5}$$