Solve 18c^2/3c+6c/3c=0

$$\frac { 18 c ^ { 2 } } { 3 c } + \frac { 6 c } { 3 c } = 0$$

Solve for c

$$c=-\frac{1}{3}\approx -0.333333333$$

Steps Using Factoring

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3c.

$$18c^{2}+6c=0$$

Factor out c.

$$c\left(18c+6\right)=0$$

To find equation solutions, solve c=0 and 18c+6=0.

$$c=0 c=-\frac{1}{3}$$

Variable c cannot be equal to 0.

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

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 3c.

$$18c^{2}+6c=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-6±\sqrt{6^{2}}}{2\times 18}$$

Take the square root of 6^{2}.

$$c=\frac{-6±6}{2\times 18}$$

Multiply 2 times 18.

$$c=\frac{-6±6}{36}$$

Now solve the equation c=\frac{-6±6}{36} when ± is plus. Add -6 to 6.

$$c=\frac{0}{36}$$

Divide 0 by 36.

$$c=0$$

Now solve the equation c=\frac{-6±6}{36} when ± is minus. Subtract 6 from -6.

$$c=-\frac{12}{36}$$

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

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

The equation is now solved.

$$c=0 c=-\frac{1}{3}$$

Variable c cannot be equal to 0.

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

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 3c.

$$18c^{2}+6c=0$$

Divide both sides by 18.

$$\frac{18c^{2}+6c}{18}=\frac{0}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$c^{2}+\frac{6}{18}c=\frac{0}{18}$$

Reduce the fraction \frac{6}{18} to lowest terms by extracting and canceling out 6.

$$c^{2}+\frac{1}{3}c=\frac{0}{18}$$

Divide 0 by 18.

$$c^{2}+\frac{1}{3}c=0$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}+\frac{1}{3}c+\left(\frac{1}{6}\right)^{2}=\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$c^{2}+\frac{1}{3}c+\frac{1}{36}=\frac{1}{36}$$

Factor c^{2}+\frac{1}{3}c+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c+\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c+\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$c+\frac{1}{6}=\frac{1}{6} c+\frac{1}{6}=-\frac{1}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$c=0 c=-\frac{1}{3}$$

Variable c cannot be equal to 0.

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