Solve -3/2x+3/12=1/2x+1/2

$$- \frac { 3 } { 2 } x + \frac { 3 } { 12 } = \frac { 1 } { 2 x } + \frac { 1 } { 2 }$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{47}i}{12}\approx -0.083333333+0.57130455i$$

$$x=\frac{-\sqrt{47}i-1}{12}\approx -0.083333333-0.57130455i$$

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 12x, the least common multiple of 2,12,2x.

$$-\frac{3}{2}x\times 12x+x\times 3=6+12x\times \frac{1}{2}$$

Multiply -\frac{3}{2} and 12 to get -18.

$$-18xx+x\times 3=6+12x\times \frac{1}{2}$$

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

$$-18x^{2}+x\times 3=6+12x\times \frac{1}{2}$$

Multiply 12 and \frac{1}{2} to get 6.

$$-18x^{2}+x\times 3=6+6x$$

Subtract 6 from both sides.

$$-18x^{2}+x\times 3-6=6x$$

Subtract 6x from both sides.

$$-18x^{2}+x\times 3-6-6x=0$$

Combine x\times 3 and -6x to get -3x.

$$-18x^{2}-3x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, -3 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-18\right)\left(-6\right)}}{2\left(-18\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-18\right)\left(-6\right)}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$x=\frac{-\left(-3\right)±\sqrt{9+72\left(-6\right)}}{2\left(-18\right)}$$

Multiply 72 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9-432}}{2\left(-18\right)}$$

Add 9 to -432.

$$x=\frac{-\left(-3\right)±\sqrt{-423}}{2\left(-18\right)}$$

Take the square root of -423.

$$x=\frac{-\left(-3\right)±3\sqrt{47}i}{2\left(-18\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{47}i}{2\left(-18\right)}$$

Multiply 2 times -18.

$$x=\frac{3±3\sqrt{47}i}{-36}$$

Now solve the equation x=\frac{3±3\sqrt{47}i}{-36} when ± is plus. Add 3 to 3i\sqrt{47}.

$$x=\frac{3+3\sqrt{47}i}{-36}$$

Divide 3+3i\sqrt{47} by -36.

$$x=\frac{-\sqrt{47}i-1}{12}$$

Now solve the equation x=\frac{3±3\sqrt{47}i}{-36} when ± is minus. Subtract 3i\sqrt{47} from 3.

$$x=\frac{-3\sqrt{47}i+3}{-36}$$

Divide 3-3i\sqrt{47} by -36.

$$x=\frac{-1+\sqrt{47}i}{12}$$

The equation is now solved.

$$x=\frac{-\sqrt{47}i-1}{12} x=\frac{-1+\sqrt{47}i}{12}$$

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 12x, the least common multiple of 2,12,2x.

$$-\frac{3}{2}x\times 12x+x\times 3=6+12x\times \frac{1}{2}$$

Multiply -\frac{3}{2} and 12 to get -18.

$$-18xx+x\times 3=6+12x\times \frac{1}{2}$$

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

$$-18x^{2}+x\times 3=6+12x\times \frac{1}{2}$$

Multiply 12 and \frac{1}{2} to get 6.

$$-18x^{2}+x\times 3=6+6x$$

Subtract 6x from both sides.

$$-18x^{2}+x\times 3-6x=6$$

Combine x\times 3 and -6x to get -3x.

$$-18x^{2}-3x=6$$

Divide both sides by -18.

$$\frac{-18x^{2}-3x}{-18}=\frac{6}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$x^{2}+\left(-\frac{3}{-18}\right)x=\frac{6}{-18}$$

Reduce the fraction \frac{-3}{-18} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{6}x=\frac{6}{-18}$$

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

$$x^{2}+\frac{1}{6}x=-\frac{1}{3}$$

Divide \frac{1}{6}, the coefficient of the x term, by 2 to get \frac{1}{12}. Then add the square of \frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{6}x+\left(\frac{1}{12}\right)^{2}=-\frac{1}{3}+\left(\frac{1}{12}\right)^{2}$$

Square \frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=-\frac{1}{3}+\frac{1}{144}$$

Add -\frac{1}{3} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=-\frac{47}{144}$$

Factor x^{2}+\frac{1}{6}x+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{12}\right)^{2}=-\frac{47}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{12}\right)^{2}}=\sqrt{-\frac{47}{144}}$$

Simplify.

$$x+\frac{1}{12}=\frac{\sqrt{47}i}{12} x+\frac{1}{12}=-\frac{\sqrt{47}i}{12}$$

Subtract \frac{1}{12} from both sides of the equation.

$$x=\frac{-1+\sqrt{47}i}{12} x=\frac{-\sqrt{47}i-1}{12}$$