Solve 3x^2+1=2x

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{2}i}{3}\approx 0.333333333+0.471404521i$$

$$x=\frac{-\sqrt{2}i+1}{3}\approx 0.333333333-0.471404521i$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

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.

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

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

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

Square -2.

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

Multiply -4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12}}{2\times 3}$$

Add 4 to -12.

$$x=\frac{-\left(-2\right)±\sqrt{-8}}{2\times 3}$$

Take the square root of -8.

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

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{2}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±2\sqrt{2}i}{6}$$

Now solve the equation x=\frac{2±2\sqrt{2}i}{6} when ± is plus. Add 2 to 2i\sqrt{2}.

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

Divide 2+2i\sqrt{2} by 6.

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

Now solve the equation x=\frac{2±2\sqrt{2}i}{6} when ± is minus. Subtract 2i\sqrt{2} from 2.

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

Divide 2-2i\sqrt{2} by 6.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 2x from both sides.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{1}{3}+\frac{1}{9}$$

Add -\frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{2}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

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