Solve 2x^2+1/2=x

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{3}i}{4}\approx 0.25+0.433012702i$$

$$x=\frac{-\sqrt{3}i+1}{4}\approx 0.25-0.433012702i$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$2x^{2}+\frac{1}{2}-x=0$$

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.

$$2x^{2}-x+\frac{1}{2}=0$$

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

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

Multiply -4 times 2.

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

Multiply -8 times \frac{1}{2}.

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

Add 1 to -4.

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

Take the square root of -3.

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

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{3}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{3}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{3}i}{4} when ± is plus. Add 1 to i\sqrt{3}.

$$x=\frac{1+\sqrt{3}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{3}i}{4} when ± is minus. Subtract i\sqrt{3} from 1.

$$x=\frac{-\sqrt{3}i+1}{4}$$

The equation is now solved.

$$x=\frac{1+\sqrt{3}i}{4} x=\frac{-\sqrt{3}i+1}{4}$$

Steps for Completing the Square

Subtract x from both sides.

$$2x^{2}+\frac{1}{2}-x=0$$

Subtract \frac{1}{2} from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{1}{2} by 2.

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

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.

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

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{1}{4}+\frac{1}{16}$$

Add -\frac{1}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{3}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{3}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{3}i}{4} x-\frac{1}{4}=-\frac{\sqrt{3}i}{4}$$

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

$$x=\frac{1+\sqrt{3}i}{4} x=\frac{-\sqrt{3}i+1}{4}$$