Solve 6+c^2/2sqrt{19c}=1/2

$$\frac { 6 + c ^ { 2 } } { 2 \sqrt { 19 } c } = \frac { 1 } { 2 }$$

Solve for c

$$c=\frac{\sqrt{19}+\sqrt{5}i}{2}\approx 2.179449472+1.118033989i$$

$$c=\frac{-\sqrt{5}i+\sqrt{19}}{2}\approx 2.179449472-1.118033989i$$

Steps Using the Quadratic Formula

Rationalize the denominator of \frac{6+c^{2}}{2\sqrt{19}c} by multiplying numerator and denominator by \sqrt{19}.

$$\frac{\left(6+c^{2}\right)\sqrt{19}}{2\left(\sqrt{19}\right)^{2}c}=\frac{1}{2}$$

The square of \sqrt{19} is 19.

$$\frac{\left(6+c^{2}\right)\sqrt{19}}{2\times 19c}=\frac{1}{2}$$

Multiply 2 and 19 to get 38.

$$\frac{\left(6+c^{2}\right)\sqrt{19}}{38c}=\frac{1}{2}$$

Use the distributive property to multiply 6+c^{2} by \sqrt{19}.

$$\frac{6\sqrt{19}+c^{2}\sqrt{19}}{38c}=\frac{1}{2}$$

Subtract \frac{1}{2} from both sides.

$$\frac{6\sqrt{19}+c^{2}\sqrt{19}}{38c}-\frac{1}{2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 38c and 2 is 38c. Multiply \frac{1}{2} times \frac{19c}{19c}.

$$\frac{6\sqrt{19}+c^{2}\sqrt{19}}{38c}-\frac{19c}{38c}=0$$

Since \frac{6\sqrt{19}+c^{2}\sqrt{19}}{38c} and \frac{19c}{38c} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6\sqrt{19}+c^{2}\sqrt{19}-19c}{38c}=0$$

Do the multiplications in 6\sqrt{19}+c^{2}\sqrt{19}-19c.

$$\frac{6\sqrt{19}+\sqrt{19}c^{2}-19c}{38c}=0$$

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 38c.

$$6\sqrt{19}+\sqrt{19}c^{2}-19c=0$$

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.

$$\sqrt{19}c^{2}-19c+6\sqrt{19}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \sqrt{19} for a, -19 for b, and 6\sqrt{19} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\sqrt{19}\times 6\sqrt{19}}}{2\sqrt{19}}$$

Square -19.

$$c=\frac{-\left(-19\right)±\sqrt{361-4\sqrt{19}\times 6\sqrt{19}}}{2\sqrt{19}}$$

Multiply -4 times \sqrt{19}.

$$c=\frac{-\left(-19\right)±\sqrt{361+\left(-4\sqrt{19}\right)\times 6\sqrt{19}}}{2\sqrt{19}}$$

Multiply -4\sqrt{19} times 6\sqrt{19}.

$$c=\frac{-\left(-19\right)±\sqrt{361-456}}{2\sqrt{19}}$$

Add 361 to -456.

$$c=\frac{-\left(-19\right)±\sqrt{-95}}{2\sqrt{19}}$$

Take the square root of -95.

$$c=\frac{-\left(-19\right)±\sqrt{95}i}{2\sqrt{19}}$$

The opposite of -19 is 19.

$$c=\frac{19±\sqrt{95}i}{2\sqrt{19}}$$

Now solve the equation c=\frac{19±\sqrt{95}i}{2\sqrt{19}} when ± is plus. Add 19 to i\sqrt{95}.

$$c=\frac{19+\sqrt{95}i}{2\sqrt{19}}$$

Divide 19+i\sqrt{95} by 2\sqrt{19}.

$$c=\frac{\sqrt{19}+\sqrt{5}i}{2}$$

Now solve the equation c=\frac{19±\sqrt{95}i}{2\sqrt{19}} when ± is minus. Subtract i\sqrt{95} from 19.

$$c=\frac{-\sqrt{95}i+19}{2\sqrt{19}}$$

Divide 19-i\sqrt{95} by 2\sqrt{19}.

$$c=\frac{-\sqrt{5}i+\sqrt{19}}{2}$$

The equation is now solved.

$$c=\frac{\sqrt{19}+\sqrt{5}i}{2} c=\frac{-\sqrt{5}i+\sqrt{19}}{2}$$

Steps for Completing the Square

Rationalize the denominator of \frac{6+c^{2}}{2\sqrt{19}c} by multiplying numerator and denominator by \sqrt{19}.

$$\frac{\left(6+c^{2}\right)\sqrt{19}}{2\left(\sqrt{19}\right)^{2}c}=\frac{1}{2}$$

The square of \sqrt{19} is 19.

$$\frac{\left(6+c^{2}\right)\sqrt{19}}{2\times 19c}=\frac{1}{2}$$

Multiply 2 and 19 to get 38.

$$\frac{\left(6+c^{2}\right)\sqrt{19}}{38c}=\frac{1}{2}$$

Use the distributive property to multiply 6+c^{2} by \sqrt{19}.

$$\frac{6\sqrt{19}+c^{2}\sqrt{19}}{38c}=\frac{1}{2}$$

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 38c, the least common multiple of 38c,2.

$$6\sqrt{19}+c^{2}\sqrt{19}=19c$$

Subtract 19c from both sides.

$$6\sqrt{19}+c^{2}\sqrt{19}-19c=0$$

Subtract 6\sqrt{19} from both sides. Anything subtracted from zero gives its negation.

$$c^{2}\sqrt{19}-19c=-6\sqrt{19}$$

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.

$$\sqrt{19}c^{2}-19c=-6\sqrt{19}$$

Divide both sides by \sqrt{19}.

$$\frac{\sqrt{19}c^{2}-19c}{\sqrt{19}}=-\frac{6\sqrt{19}}{\sqrt{19}}$$

Dividing by \sqrt{19} undoes the multiplication by \sqrt{19}.

$$c^{2}+\left(-\frac{19}{\sqrt{19}}\right)c=-\frac{6\sqrt{19}}{\sqrt{19}}$$

Divide -19 by \sqrt{19}.

$$c^{2}+\left(-\sqrt{19}\right)c=-\frac{6\sqrt{19}}{\sqrt{19}}$$

Divide -6\sqrt{19} by \sqrt{19}.

$$c^{2}+\left(-\sqrt{19}\right)c=-6$$

Divide -\sqrt{19}, the coefficient of the x term, by 2 to get -\frac{\sqrt{19}}{2}. Then add the square of -\frac{\sqrt{19}}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}+\left(-\sqrt{19}\right)c+\left(-\frac{\sqrt{19}}{2}\right)^{2}=-6+\left(-\frac{\sqrt{19}}{2}\right)^{2}$$

Square -\frac{\sqrt{19}}{2}.

$$c^{2}+\left(-\sqrt{19}\right)c+\frac{19}{4}=-6+\frac{19}{4}$$

Add -6 to \frac{19}{4}.

$$c^{2}+\left(-\sqrt{19}\right)c+\frac{19}{4}=-\frac{5}{4}$$

Factor c^{2}+\left(-\sqrt{19}\right)c+\frac{19}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c-\frac{\sqrt{19}}{2}\right)^{2}=-\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-\frac{\sqrt{19}}{2}\right)^{2}}=\sqrt{-\frac{5}{4}}$$

Simplify.

$$c-\frac{\sqrt{19}}{2}=\frac{\sqrt{5}i}{2} c-\frac{\sqrt{19}}{2}=-\frac{\sqrt{5}i}{2}$$

Add \frac{\sqrt{19}}{2} to both sides of the equation.

$$c=\frac{\sqrt{19}+\sqrt{5}i}{2} c=\frac{-\sqrt{5}i+\sqrt{19}}{2}$$