Solve 2x^2+1/2x+1=x(x-2)

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

Solve for x

$$x=-2$$

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

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x-2.

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

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

Add 2x to both sides.

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

Combine \frac{1}{2}x and 2x to get \frac{5}{2}x.

$$x^{2}+\frac{5}{2}x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, \frac{5}{2} for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{5}{2}±\sqrt{\left(\frac{5}{2}\right)^{2}-4}}{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-4}}{2}$$

Add \frac{25}{4} to -4.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{9}{4}}}{2}$$

Take the square root of \frac{9}{4}.

$$x=\frac{-\frac{5}{2}±\frac{3}{2}}{2}$$

Now solve the equation x=\frac{-\frac{5}{2}±\frac{3}{2}}{2} when ± is plus. Add -\frac{5}{2} to \frac{3}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Now solve the equation x=\frac{-\frac{5}{2}±\frac{3}{2}}{2} when ± is minus. Subtract \frac{3}{2} from -\frac{5}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply x by x-2.

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

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

Add 2x to both sides.

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

Combine \frac{1}{2}x and 2x to get \frac{5}{2}x.

$$x^{2}+\frac{5}{2}x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+\frac{5}{2}x=-1$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=-1+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-1+\frac{25}{16}$$

Add -1 to \frac{25}{16}.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{9}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{3}{4} x+\frac{5}{4}=-\frac{3}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

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