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

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

Solve for x

$$x = \frac{\sqrt{17} - 1}{2} \approx 1.561552813$$

$$x=\frac{-\sqrt{17}-1}{2}\approx -2.561552813$$

Steps Using the Quadratic Formula

Combine 2x and -x to get x.

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

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

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

$$x^{2}+x-4=0$$

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

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

Square 1.

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

Multiply -4 times -4.

$$x=\frac{-1±\sqrt{1+16}}{2}$$

Add 1 to 16.

$$x=\frac{-1±\sqrt{17}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{17}}{2} when ± is plus. Add -1 to \sqrt{17}.

$$x=\frac{\sqrt{17}-1}{2}$$

Now solve the equation x=\frac{-1±\sqrt{17}}{2} when ± is minus. Subtract \sqrt{17} from -1.

$$x=\frac{-\sqrt{17}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-1}{2} x=\frac{-\sqrt{17}-1}{2}$$

Steps for Completing the Square

Combine 2x and -x to get x.

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

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

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

Add 2 to both sides.

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

Add 2 and 2 to get 4.

$$x^{2}+x=4$$

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.

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

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

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

Add 4 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{17}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{17}}{2} x+\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

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

$$x=\frac{\sqrt{17}-1}{2} x=\frac{-\sqrt{17}-1}{2}$$