Solve 5/7=c^2+2c+1+1/2-(c^2+2c+1)

$$\frac { 5 } { 7 } = \frac { c ^ { 2 } + 2 c + 1 + 1 } { 2 - ( c ^ { 2 } + 2 c + 1 ) }$$

Solve for c

$$c = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$c=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 7\left(c-\left(\sqrt{2}-1\right)\right)\left(c-\left(-\sqrt{2}-1\right)\right), the least common multiple of 7,2-\left(c^{2}+2c+1\right).

$$5\left(c-\left(\sqrt{2}-1\right)\right)\left(c-\left(-\sqrt{2}-1\right)\right)=-7\left(c^{2}+2c+1+1\right)$$

To find the opposite of \sqrt{2}-1, find the opposite of each term.

$$5\left(c-\sqrt{2}+1\right)\left(c-\left(-\sqrt{2}-1\right)\right)=-7\left(c^{2}+2c+1+1\right)$$

To find the opposite of -\sqrt{2}-1, find the opposite of each term.

$$5\left(c-\sqrt{2}+1\right)\left(c+\sqrt{2}+1\right)=-7\left(c^{2}+2c+1+1\right)$$

Use the distributive property to multiply 5 by c-\sqrt{2}+1.

$$\left(5c-5\sqrt{2}+5\right)\left(c+\sqrt{2}+1\right)=-7\left(c^{2}+2c+1+1\right)$$

Use the distributive property to multiply 5c-5\sqrt{2}+5 by c+\sqrt{2}+1 and combine like terms.

$$5c^{2}+10c-5\left(\sqrt{2}\right)^{2}+5=-7\left(c^{2}+2c+1+1\right)$$

The square of \sqrt{2} is 2.

$$5c^{2}+10c-5\times 2+5=-7\left(c^{2}+2c+1+1\right)$$

Multiply -5 and 2 to get -10.

$$5c^{2}+10c-10+5=-7\left(c^{2}+2c+1+1\right)$$

Add -10 and 5 to get -5.

$$5c^{2}+10c-5=-7\left(c^{2}+2c+1+1\right)$$

Add 1 and 1 to get 2.

$$5c^{2}+10c-5=-7\left(c^{2}+2c+2\right)$$

Use the distributive property to multiply -7 by c^{2}+2c+2.

$$5c^{2}+10c-5=-7c^{2}-14c-14$$

Add 7c^{2} to both sides.

$$5c^{2}+10c-5+7c^{2}=-14c-14$$

Combine 5c^{2} and 7c^{2} to get 12c^{2}.

$$12c^{2}+10c-5=-14c-14$$

Add 14c to both sides.

$$12c^{2}+10c-5+14c=-14$$

Combine 10c and 14c to get 24c.

$$12c^{2}+24c-5=-14$$

Add 14 to both sides.

$$12c^{2}+24c-5+14=0$$

Add -5 and 14 to get 9.

$$12c^{2}+24c+9=0$$

Divide both sides by 3.

$$4c^{2}+8c+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4c^{2}+ac+bc+3. To find a and b, set up a system to be solved.

$$a+b=8 ab=4\times 3=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=2 b=6$$

Rewrite 4c^{2}+8c+3 as \left(4c^{2}+2c\right)+\left(6c+3\right).

$$\left(4c^{2}+2c\right)+\left(6c+3\right)$$

Factor out 2c in the first and 3 in the second group.

$$2c\left(2c+1\right)+3\left(2c+1\right)$$

Factor out common term 2c+1 by using distributive property.

$$\left(2c+1\right)\left(2c+3\right)$$

To find equation solutions, solve 2c+1=0 and 2c+3=0.

$$c=-\frac{1}{2} c=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 7\left(c-\left(\sqrt{2}-1\right)\right)\left(c-\left(-\sqrt{2}-1\right)\right), the least common multiple of 7,2-\left(c^{2}+2c+1\right).

$$5\left(c-\left(\sqrt{2}-1\right)\right)\left(c-\left(-\sqrt{2}-1\right)\right)=-7\left(c^{2}+2c+1+1\right)$$

To find the opposite of \sqrt{2}-1, find the opposite of each term.

$$5\left(c-\sqrt{2}+1\right)\left(c-\left(-\sqrt{2}-1\right)\right)=-7\left(c^{2}+2c+1+1\right)$$

To find the opposite of -\sqrt{2}-1, find the opposite of each term.

$$5\left(c-\sqrt{2}+1\right)\left(c+\sqrt{2}+1\right)=-7\left(c^{2}+2c+1+1\right)$$

Use the distributive property to multiply 5 by c-\sqrt{2}+1.

$$\left(5c-5\sqrt{2}+5\right)\left(c+\sqrt{2}+1\right)=-7\left(c^{2}+2c+1+1\right)$$

Use the distributive property to multiply 5c-5\sqrt{2}+5 by c+\sqrt{2}+1 and combine like terms.

$$5c^{2}+10c-5\left(\sqrt{2}\right)^{2}+5=-7\left(c^{2}+2c+1+1\right)$$

The square of \sqrt{2} is 2.

$$5c^{2}+10c-5\times 2+5=-7\left(c^{2}+2c+1+1\right)$$

Multiply -5 and 2 to get -10.

$$5c^{2}+10c-10+5=-7\left(c^{2}+2c+1+1\right)$$

Add -10 and 5 to get -5.

$$5c^{2}+10c-5=-7\left(c^{2}+2c+1+1\right)$$

Add 1 and 1 to get 2.

$$5c^{2}+10c-5=-7\left(c^{2}+2c+2\right)$$

Use the distributive property to multiply -7 by c^{2}+2c+2.

$$5c^{2}+10c-5=-7c^{2}-14c-14$$

Add 7c^{2} to both sides.

$$5c^{2}+10c-5+7c^{2}=-14c-14$$

Combine 5c^{2} and 7c^{2} to get 12c^{2}.

$$12c^{2}+10c-5=-14c-14$$

Add 14c to both sides.

$$12c^{2}+10c-5+14c=-14$$

Combine 10c and 14c to get 24c.

$$12c^{2}+24c-5=-14$$

Add 14 to both sides.

$$12c^{2}+24c-5+14=0$$

Add -5 and 14 to get 9.

$$12c^{2}+24c+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 24 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-24±\sqrt{24^{2}-4\times 12\times 9}}{2\times 12}$$

Square 24.

$$c=\frac{-24±\sqrt{576-4\times 12\times 9}}{2\times 12}$$

Multiply -4 times 12.

$$c=\frac{-24±\sqrt{576-48\times 9}}{2\times 12}$$

Multiply -48 times 9.

$$c=\frac{-24±\sqrt{576-432}}{2\times 12}$$

Add 576 to -432.

$$c=\frac{-24±\sqrt{144}}{2\times 12}$$

Take the square root of 144.

$$c=\frac{-24±12}{2\times 12}$$

Multiply 2 times 12.

$$c=\frac{-24±12}{24}$$

Now solve the equation c=\frac{-24±12}{24} when ± is plus. Add -24 to 12.

$$c=-\frac{12}{24}$$

Reduce the fraction \frac{-12}{24} to lowest terms by extracting and canceling out 12.

$$c=-\frac{1}{2}$$

Now solve the equation c=\frac{-24±12}{24} when ± is minus. Subtract 12 from -24.

$$c=-\frac{36}{24}$$

Reduce the fraction \frac{-36}{24} to lowest terms by extracting and canceling out 12.

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

The equation is now solved.

$$c=-\frac{1}{2} c=-\frac{3}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 7\left(c-\left(\sqrt{2}-1\right)\right)\left(c-\left(-\sqrt{2}-1\right)\right), the least common multiple of 7,2-\left(c^{2}+2c+1\right).

$$5\left(c-\left(\sqrt{2}-1\right)\right)\left(c-\left(-\sqrt{2}-1\right)\right)=-7\left(c^{2}+2c+1+1\right)$$

To find the opposite of \sqrt{2}-1, find the opposite of each term.

$$5\left(c-\sqrt{2}+1\right)\left(c-\left(-\sqrt{2}-1\right)\right)=-7\left(c^{2}+2c+1+1\right)$$

To find the opposite of -\sqrt{2}-1, find the opposite of each term.

$$5\left(c-\sqrt{2}+1\right)\left(c+\sqrt{2}+1\right)=-7\left(c^{2}+2c+1+1\right)$$

Use the distributive property to multiply 5 by c-\sqrt{2}+1.

$$\left(5c-5\sqrt{2}+5\right)\left(c+\sqrt{2}+1\right)=-7\left(c^{2}+2c+1+1\right)$$

Use the distributive property to multiply 5c-5\sqrt{2}+5 by c+\sqrt{2}+1 and combine like terms.

$$5c^{2}+10c-5\left(\sqrt{2}\right)^{2}+5=-7\left(c^{2}+2c+1+1\right)$$

The square of \sqrt{2} is 2.

$$5c^{2}+10c-5\times 2+5=-7\left(c^{2}+2c+1+1\right)$$

Multiply -5 and 2 to get -10.

$$5c^{2}+10c-10+5=-7\left(c^{2}+2c+1+1\right)$$

Add -10 and 5 to get -5.

$$5c^{2}+10c-5=-7\left(c^{2}+2c+1+1\right)$$

Add 1 and 1 to get 2.

$$5c^{2}+10c-5=-7\left(c^{2}+2c+2\right)$$

Use the distributive property to multiply -7 by c^{2}+2c+2.

$$5c^{2}+10c-5=-7c^{2}-14c-14$$

Add 7c^{2} to both sides.

$$5c^{2}+10c-5+7c^{2}=-14c-14$$

Combine 5c^{2} and 7c^{2} to get 12c^{2}.

$$12c^{2}+10c-5=-14c-14$$

Add 14c to both sides.

$$12c^{2}+10c-5+14c=-14$$

Combine 10c and 14c to get 24c.

$$12c^{2}+24c-5=-14$$

Add 5 to both sides.

$$12c^{2}+24c=-14+5$$

Add -14 and 5 to get -9.

$$12c^{2}+24c=-9$$

Divide both sides by 12.

$$\frac{12c^{2}+24c}{12}=-\frac{9}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$c^{2}+\frac{24}{12}c=-\frac{9}{12}$$

Divide 24 by 12.

$$c^{2}+2c=-\frac{9}{12}$$

Reduce the fraction \frac{-9}{12} to lowest terms by extracting and canceling out 3.

$$c^{2}+2c=-\frac{3}{4}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}+2c+1^{2}=-\frac{3}{4}+1^{2}$$

Square 1.

$$c^{2}+2c+1=-\frac{3}{4}+1$$

Add -\frac{3}{4} to 1.

$$c^{2}+2c+1=\frac{1}{4}$$

Factor c^{2}+2c+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(c+1\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$c+1=\frac{1}{2} c+1=-\frac{1}{2}$$

Subtract 1 from both sides of the equation.

$$c=-\frac{1}{2} c=-\frac{3}{2}$$