Solve c^2-15/14c=-1/7

$$\frac { c ^ { 2 } - 15 } { 14 c } = \frac { - 1 } { 7 }$$

Solve for c

$$c=-5$$

$$c=3$$

Steps Using Factoring

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 14c, the least common multiple of 14c,7.

$$c^{2}-15=2c\left(-1\right)$$

Multiply 2 and -1 to get -2.

$$c^{2}-15=-2c$$

Add 2c to both sides.

$$c^{2}-15+2c=0$$

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

$$c^{2}+2c-15=0$$

To solve the equation, factor c^{2}+2c-15 using formula c^{2}+\left(a+b\right)c+ab=\left(c+a\right)\left(c+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite factored expression \left(c+a\right)\left(c+b\right) using the obtained values.

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

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

$$c=3 c=-5$$

Steps Using Factoring By Grouping

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 14c, the least common multiple of 14c,7.

$$c^{2}-15=2c\left(-1\right)$$

Multiply 2 and -1 to get -2.

$$c^{2}-15=-2c$$

Add 2c to both sides.

$$c^{2}-15+2c=0$$

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

$$c^{2}+2c-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as c^{2}+ac+bc-15. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-15\right)=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite c^{2}+2c-15 as \left(c^{2}-3c\right)+\left(5c-15\right).

$$\left(c^{2}-3c\right)+\left(5c-15\right)$$

Factor out c in the first and 5 in the second group.

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

Factor out common term c-3 by using distributive property.

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

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

$$c=3 c=-5$$

Steps Using the Quadratic Formula

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 14c, the least common multiple of 14c,7.

$$c^{2}-15=2c\left(-1\right)$$

Multiply 2 and -1 to get -2.

$$c^{2}-15=-2c$$

Add 2c to both sides.

$$c^{2}-15+2c=0$$

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.

$$c^{2}+2c-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-2±\sqrt{2^{2}-4\left(-15\right)}}{2}$$

Square 2.

$$c=\frac{-2±\sqrt{4-4\left(-15\right)}}{2}$$

Multiply -4 times -15.

$$c=\frac{-2±\sqrt{4+60}}{2}$$

Add 4 to 60.

$$c=\frac{-2±\sqrt{64}}{2}$$

Take the square root of 64.

$$c=\frac{-2±8}{2}$$

Now solve the equation c=\frac{-2±8}{2} when ± is plus. Add -2 to 8.

$$c=\frac{6}{2}$$

Divide 6 by 2.

$$c=3$$

Now solve the equation c=\frac{-2±8}{2} when ± is minus. Subtract 8 from -2.

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

Divide -10 by 2.

$$c=-5$$

The equation is now solved.

$$c=3 c=-5$$

Steps for Completing the Square

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 14c, the least common multiple of 14c,7.

$$c^{2}-15=2c\left(-1\right)$$

Multiply 2 and -1 to get -2.

$$c^{2}-15=-2c$$

Add 2c to both sides.

$$c^{2}-15+2c=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$c^{2}+2c=15$$

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.

$$c^{2}+2c+1^{2}=15+1^{2}$$

Square 1.

$$c^{2}+2c+1=15+1$$

Add 15 to 1.

$$c^{2}+2c+1=16$$

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}.

$$\left(c+1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c+1\right)^{2}}=\sqrt{16}$$

Simplify.

$$c+1=4 c+1=-4$$

Subtract 1 from both sides of the equation.

$$c=3 c=-5$$