Solve -x^2+2=x-2

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

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

Subtract x from both sides.

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

Add 2 to both sides.

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

Add 2 and 2 to get 4.

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

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.

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

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}.

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

Multiply -4 times -1.

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

Multiply 4 times 4.

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

Add 1 to 16.

$$x=\frac{-\left(-1\right)±\sqrt{17}}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{17}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±\sqrt{17}}{-2}$$

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

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

Divide 1+\sqrt{17} by -2.

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

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

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

Divide 1-\sqrt{17} by -2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x from both sides.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide -4 by -1.

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

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.

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