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

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

Solve for x

$$x=-2$$

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Express 2\left(-\frac{x}{2}\right) as a single fraction.

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

Cancel out 2 and 2.

$$-x\left(1-2x\right)+2x=2-2x$$

Use the distributive property to multiply -x by 1-2x.

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

Combine -x and 2x to get x.

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

Subtract 2 from both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

$$3x+2x^{2}-2=0$$

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.

$$2x^{2}+3x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-2\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-3±\sqrt{9+16}}{2\times 2}$$

Add 9 to 16.

$$x=\frac{-3±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-3±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±5}{4}$$

Now solve the equation x=\frac{-3±5}{4} when ± is plus. Add -3 to 5.

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

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

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

Now solve the equation x=\frac{-3±5}{4} when ± is minus. Subtract 5 from -3.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Express 2\left(-\frac{x}{2}\right) as a single fraction.

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

Cancel out 2 and 2.

$$-x\left(1-2x\right)+2x=2-2x$$

Use the distributive property to multiply -x by 1-2x.

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

Combine -x and 2x to get x.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

$$3x+2x^{2}=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}+3x=2$$

Divide both sides by 2.

$$\frac{2x^{2}+3x}{2}=\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{3}{2}x=\frac{2}{2}$$

Divide 2 by 2.

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

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{25}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

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

Subtract \frac{3}{4} from both sides of the equation.

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