Solve (x)=(3x^2-2x+1)

$$( x ) = ( 3 x ^ { 2 } - 2 x + 1 )$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{3}i}{6}+\frac{1}{2}\approx 0.5-0.288675135i$$

$$x=\frac{\sqrt{3}i}{6}+\frac{1}{2}\approx 0.5+0.288675135i$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$x-3x^{2}=-2x+1$$

Add 2x to both sides.

$$x-3x^{2}+2x=1$$

Combine x and 2x to get 3x.

$$3x-3x^{2}=1$$

Subtract 1 from both sides.

$$3x-3x^{2}-1=0$$

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.

$$-3x^{2}+3x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-3±\sqrt{9+12\left(-1\right)}}{2\left(-3\right)}$$

Multiply 12 times -1.

$$x=\frac{-3±\sqrt{9-12}}{2\left(-3\right)}$$

Add 9 to -12.

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

Take the square root of -3.

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

Multiply 2 times -3.

$$x=\frac{-3±\sqrt{3}i}{-6}$$

Now solve the equation x=\frac{-3±\sqrt{3}i}{-6} when ± is plus. Add -3 to i\sqrt{3}.

$$x=\frac{-3+\sqrt{3}i}{-6}$$

Divide -3+i\sqrt{3} by -6.

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

Now solve the equation x=\frac{-3±\sqrt{3}i}{-6} when ± is minus. Subtract i\sqrt{3} from -3.

$$x=\frac{-\sqrt{3}i-3}{-6}$$

Divide -3-i\sqrt{3} by -6.

$$x=\frac{\sqrt{3}i}{6}+\frac{1}{2}$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$x-3x^{2}=-2x+1$$

Add 2x to both sides.

$$x-3x^{2}+2x=1$$

Combine x and 2x to get 3x.

$$3x-3x^{2}=1$$

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.

$$-3x^{2}+3x=1$$

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 3 by -3.

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

Divide 1 by -3.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-\frac{1}{3}+\frac{1}{4}$$

Add -\frac{1}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=-\frac{1}{12}$$

Factor x^{2}-x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{3}i}{6} x-\frac{1}{2}=-\frac{\sqrt{3}i}{6}$$

Add \frac{1}{2} to both sides of the equation.

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