Solve 2x^2+1/2=2x

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

Solve for x

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

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.

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

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

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

Square -2.

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

Multiply -4 times 2.

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

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

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

Add 4 to -4.

$$x=\frac{-\left(-2\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$x=-\frac{-2}{2\times 2}$$

The opposite of -2 is 2.

$$x=\frac{2}{2\times 2}$$

Multiply 2 times 2.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Steps for Completing the Square

Subtract 2x from both sides.

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

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

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

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

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

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

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

$$x^{2}-x+\frac{1}{4}=0$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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