Solve 2x^2=1-2x

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

Solve for x

$$x=\frac{\sqrt{3}-1}{2}\approx 0.366025404$$

$$x=\frac{-\sqrt{3}-1}{2}\approx -1.366025404$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

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

Add 2x to both sides.

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

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.

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

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

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

Square 2.

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

Multiply -4 times 2.

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

Multiply -8 times -1.

$$x=\frac{-2±\sqrt{4+8}}{2\times 2}$$

Add 4 to 8.

$$x=\frac{-2±\sqrt{12}}{2\times 2}$$

Take the square root of 12.

$$x=\frac{-2±2\sqrt{3}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±2\sqrt{3}}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{3}}{4} when ± is plus. Add -2 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-2}{4}$$

Divide -2+2\sqrt{3} by 4.

$$x=\frac{\sqrt{3}-1}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{3}}{4} when ± is minus. Subtract 2\sqrt{3} from -2.

$$x=\frac{-2\sqrt{3}-2}{4}$$

Divide -2-2\sqrt{3} by 4.

$$x=\frac{-\sqrt{3}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{3}-1}{2} x=\frac{-\sqrt{3}-1}{2}$$

Steps for Completing the Square

Add 2x to both sides.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 2 by 2.

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

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.

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

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

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

Add \frac{1}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{3}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{3}-1}{2} x=\frac{-\sqrt{3}-1}{2}$$