Solve 2x/3+6/x+5/3=0

$$2 \frac { x } { 3 } + \frac { 6 } { x } + \frac { 5 } { 3 } = 0$$

Solve for x (complex solution)

$$x=\frac{-5+\sqrt{119}i}{4}\approx -1.25+2.727178029i$$

$$x=\frac{-\sqrt{119}i-5}{4}\approx -1.25-2.727178029i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x, the least common multiple of 3,x.

$$2xx+3\times 6+3x\times \frac{5}{3}=0$$

Multiply x and x to get x^{2}.

$$2x^{2}+3\times 6+3x\times \frac{5}{3}=0$$

Multiply 3 and 6 to get 18.

$$2x^{2}+18+3x\times \frac{5}{3}=0$$

Cancel out 3 and 3.

$$2x^{2}+18+5x=0$$

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.

$$2x^{2}+5x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 18}}{2\times 2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 2\times 18}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-5±\sqrt{25-8\times 18}}{2\times 2}$$

Multiply -8 times 18.

$$x=\frac{-5±\sqrt{25-144}}{2\times 2}$$

Add 25 to -144.

$$x=\frac{-5±\sqrt{-119}}{2\times 2}$$

Take the square root of -119.

$$x=\frac{-5±\sqrt{119}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-5±\sqrt{119}i}{4}$$

Now solve the equation x=\frac{-5±\sqrt{119}i}{4} when ± is plus. Add -5 to i\sqrt{119}.

$$x=\frac{-5+\sqrt{119}i}{4}$$

Now solve the equation x=\frac{-5±\sqrt{119}i}{4} when ± is minus. Subtract i\sqrt{119} from -5.

$$x=\frac{-\sqrt{119}i-5}{4}$$

The equation is now solved.

$$x=\frac{-5+\sqrt{119}i}{4} x=\frac{-\sqrt{119}i-5}{4}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x, the least common multiple of 3,x.

$$2xx+3\times 6+3x\times \frac{5}{3}=0$$

Multiply x and x to get x^{2}.

$$2x^{2}+3\times 6+3x\times \frac{5}{3}=0$$

Multiply 3 and 6 to get 18.

$$2x^{2}+18+3x\times \frac{5}{3}=0$$

Cancel out 3 and 3.

$$2x^{2}+18+5x=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+5x=-18$$

Divide both sides by 2.

$$\frac{2x^{2}+5x}{2}=-\frac{18}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{5}{2}x=-\frac{18}{2}$$

Divide -18 by 2.

$$x^{2}+\frac{5}{2}x=-9$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=-9+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-9+\frac{25}{16}$$

Add -9 to \frac{25}{16}.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{119}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{4}\right)^{2}=-\frac{119}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{-\frac{119}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{\sqrt{119}i}{4} x+\frac{5}{4}=-\frac{\sqrt{119}i}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$x=\frac{-5+\sqrt{119}i}{4} x=\frac{-\sqrt{119}i-5}{4}$$