Solve 2x^2=x-4

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{31}i}{4}\approx 0.25+1.391941091i$$

$$x=\frac{-\sqrt{31}i+1}{4}\approx 0.25-1.391941091i$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Add 4 to both sides.

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

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

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

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-\left(-1\right)±\sqrt{1-32}}{2\times 2}$$

Add 1 to -32.

$$x=\frac{-\left(-1\right)±\sqrt{-31}}{2\times 2}$$

Take the square root of -31.

$$x=\frac{-\left(-1\right)±\sqrt{31}i}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{31}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{31}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{31}i}{4} when ± is plus. Add 1 to i\sqrt{31}.

$$x=\frac{1+\sqrt{31}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{31}i}{4} when ± is minus. Subtract i\sqrt{31} from 1.

$$x=\frac{-\sqrt{31}i+1}{4}$$

The equation is now solved.

$$x=\frac{1+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i+1}{4}$$

Steps for Completing the Square

Subtract x from both sides.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -4 by 2.

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

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.

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

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

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

Add -2 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{31}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{31}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{31}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{31}i}{4} x-\frac{1}{4}=-\frac{\sqrt{31}i}{4}$$

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

$$x=\frac{1+\sqrt{31}i}{4} x=\frac{-\sqrt{31}i+1}{4}$$