Solve x=2x^2+2x

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

Solve for x

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

$$x=0$$

Steps Using Factoring

Subtract 2x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Factor out x.

$$x\left(-1-2x\right)=0$$

To find equation solutions, solve x=0 and -1-2x=0.

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

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

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.

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

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

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{1±1}{-4}$$

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

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

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{1±1}{-4} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -1 by -2.

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

Divide 0 by -2.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{4} from both sides of the equation.

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