Solve 3/2:(1-x)+2=3x

$$\frac { 3 } { 2 } : ( 1 - x ) + 2 = 3 x$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{17}i}{6}\approx 0.833333333+0.687184271i$$

$$x=\frac{-\sqrt{17}i+5}{6}\approx 0.833333333-0.687184271i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

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

Use the distributive property to multiply -x+1 by 2.

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

Add \frac{3}{2} and 2 to get \frac{7}{2}.

$$\frac{7}{2}-2x=3x\left(-x+1\right)$$

Use the distributive property to multiply 3x by -x+1.

$$\frac{7}{2}-2x=-3x^{2}+3x$$

Add 3x^{2} to both sides.

$$\frac{7}{2}-2x+3x^{2}=3x$$

Subtract 3x from both sides.

$$\frac{7}{2}-2x+3x^{2}-3x=0$$

Combine -2x and -3x to get -5x.

$$\frac{7}{2}-5x+3x^{2}=0$$

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.

$$3x^{2}-5x+\frac{7}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -5 for b, and \frac{7}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\times \frac{7}{2}}}{2\times 3}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 3\times \frac{7}{2}}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-5\right)±\sqrt{25-12\times \frac{7}{2}}}{2\times 3}$$

Multiply -12 times \frac{7}{2}.

$$x=\frac{-\left(-5\right)±\sqrt{25-42}}{2\times 3}$$

Add 25 to -42.

$$x=\frac{-\left(-5\right)±\sqrt{-17}}{2\times 3}$$

Take the square root of -17.

$$x=\frac{-\left(-5\right)±\sqrt{17}i}{2\times 3}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{17}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{5±\sqrt{17}i}{6}$$

Now solve the equation x=\frac{5±\sqrt{17}i}{6} when ± is plus. Add 5 to i\sqrt{17}.

$$x=\frac{5+\sqrt{17}i}{6}$$

Now solve the equation x=\frac{5±\sqrt{17}i}{6} when ± is minus. Subtract i\sqrt{17} from 5.

$$x=\frac{-\sqrt{17}i+5}{6}$$

The equation is now solved.

$$x=\frac{5+\sqrt{17}i}{6} x=\frac{-\sqrt{17}i+5}{6}$$

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

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

Use the distributive property to multiply -x+1 by 2.

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

Add \frac{3}{2} and 2 to get \frac{7}{2}.

$$\frac{7}{2}-2x=3x\left(-x+1\right)$$

Use the distributive property to multiply 3x by -x+1.

$$\frac{7}{2}-2x=-3x^{2}+3x$$

Add 3x^{2} to both sides.

$$\frac{7}{2}-2x+3x^{2}=3x$$

Subtract 3x from both sides.

$$\frac{7}{2}-2x+3x^{2}-3x=0$$

Combine -2x and -3x to get -5x.

$$\frac{7}{2}-5x+3x^{2}=0$$

Subtract \frac{7}{2} from both sides. Anything subtracted from zero gives its negation.

$$-5x+3x^{2}=-\frac{7}{2}$$

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.

$$3x^{2}-5x=-\frac{7}{2}$$

Divide both sides by 3.

$$\frac{3x^{2}-5x}{3}=-\frac{\frac{7}{2}}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{5}{3}x=-\frac{\frac{7}{2}}{3}$$

Divide -\frac{7}{2} by 3.

$$x^{2}-\frac{5}{3}x=-\frac{7}{6}$$

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{7}{6}+\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{7}{6}+\frac{25}{36}$$

Add -\frac{7}{6} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{17}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{6}\right)^{2}=-\frac{17}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{-\frac{17}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{\sqrt{17}i}{6} x-\frac{5}{6}=-\frac{\sqrt{17}i}{6}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{5+\sqrt{17}i}{6} x=\frac{-\sqrt{17}i+5}{6}$$