Solve 2x=2-x^2

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

Solve for x (complex solution)

$$x=\sqrt{3}-1\approx 0.732050808$$

$$x=-\left(\sqrt{3}+1\right)\approx -2.732050808$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

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

Add x^{2} to both sides.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -2.

$$x=\frac{-2±\sqrt{4+8}}{2}$$

Add 4 to 8.

$$x=\frac{-2±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-2±2\sqrt{3}}{2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Add x^{2} to both sides.

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

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.

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

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

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

Square 1.

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

Add 2 to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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

Solve for x

$$x=\sqrt{3}-1\approx 0.732050808$$

$$x=-\sqrt{3}-1\approx -2.732050808$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

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

Add x^{2} to both sides.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -2.

$$x=\frac{-2±\sqrt{4+8}}{2}$$

Add 4 to 8.

$$x=\frac{-2±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-2±2\sqrt{3}}{2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Add x^{2} to both sides.

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

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.

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

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

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

Square 1.

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

Add 2 to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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