Solve 2=x^2+2x-1+2

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

Solve for x (complex solution)

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\left(\sqrt{2}+1\right)\approx -2.414213562$$

Steps Using the Quadratic Formula

Add -1 and 2 to get 1.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides.

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

Subtract 2 from 1 to get -1.

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

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

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

Square 2.

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

Multiply -4 times -1.

$$x=\frac{-2±\sqrt{4+4}}{2}$$

Add 4 to 4.

$$x=\frac{-2±\sqrt{8}}{2}$$

Take the square root of 8.

$$x=\frac{-2±2\sqrt{2}}{2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Add -1 and 2 to get 1.

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

Swap sides so that all variable terms are on the left hand side.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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

Solve for x

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\sqrt{2}-1\approx -2.414213562$$

Steps Using the Quadratic Formula

Add -1 and 2 to get 1.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides.

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

Subtract 2 from 1 to get -1.

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

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

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

Square 2.

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

Multiply -4 times -1.

$$x=\frac{-2±\sqrt{4+4}}{2}$$

Add 4 to 4.

$$x=\frac{-2±\sqrt{8}}{2}$$

Take the square root of 8.

$$x=\frac{-2±2\sqrt{2}}{2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Add -1 and 2 to get 1.

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

Swap sides so that all variable terms are on the left hand side.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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