Solve 2x^2+6=x

$$2 { x }^{ 2 } +6=x$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{47}i}{4}\approx 0.25+1.71391365i$$

$$x=\frac{-\sqrt{47}i+1}{4}\approx 0.25-1.71391365i$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1-48}}{2\times 2}$$

Add 1 to -48.

$$x=\frac{-\left(-1\right)±\sqrt{-47}}{2\times 2}$$

Take the square root of -47.

$$x=\frac{-\left(-1\right)±\sqrt{47}i}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{47}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{47}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{47}i}{4} when ± is plus. Add 1 to i\sqrt{47}.

$$x=\frac{1+\sqrt{47}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{47}i}{4} when ± is minus. Subtract i\sqrt{47} from 1.

$$x=\frac{-\sqrt{47}i+1}{4}$$

The equation is now solved.

$$x=\frac{1+\sqrt{47}i}{4} x=\frac{-\sqrt{47}i+1}{4}$$

Steps for Completing the Square

Subtract x from both sides.

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

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

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -6 by 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-3+\frac{1}{16}$$

Add -3 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{47}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{47}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{47}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{47}i}{4} x-\frac{1}{4}=-\frac{\sqrt{47}i}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1+\sqrt{47}i}{4} x=\frac{-\sqrt{47}i+1}{4}$$