Solve 0=2x^2-4x+6

$$0=2 { x }^{ 2 } -4x+6$$

Solve for x (complex solution)

$$x=1+\sqrt{2}i\approx 1+1.414213562i$$

$$x=-\sqrt{2}i+1\approx 1-1.414213562i$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-4x+6=0$$

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

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-\left(-4\right)±\sqrt{16-48}}{2\times 2}$$

Add 16 to -48.

$$x=\frac{-\left(-4\right)±\sqrt{-32}}{2\times 2}$$

Take the square root of -32.

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

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{2}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±4\sqrt{2}i}{4}$$

Now solve the equation x=\frac{4±4\sqrt{2}i}{4} when ± is plus. Add 4 to 4i\sqrt{2}.

$$x=\frac{4+4\sqrt{2}i}{4}$$

Divide 4+4i\sqrt{2} by 4.

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

Now solve the equation x=\frac{4±4\sqrt{2}i}{4} when ± is minus. Subtract 4i\sqrt{2} from 4.

$$x=\frac{-4\sqrt{2}i+4}{4}$$

Divide 4-4i\sqrt{2} by 4.

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

The equation is now solved.

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

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}-4x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-4x=-6$$

Divide both sides by 2.

$$\frac{2x^{2}-4x}{2}=-\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{4}{2}\right)x=-\frac{6}{2}$$

Divide -4 by 2.

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

Divide -6 by 2.

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

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

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

Add -3 to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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