Solve -x^2=-x-1/2

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

Solve for x

$$x = \frac{\sqrt{3} + 1}{2} \approx 1.366025404$$

$$x=\frac{1-\sqrt{3}}{2}\approx -0.366025404$$

Steps Using the Quadratic Formula

Add x to both sides.

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

Add \frac{1}{2} to both sides.

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

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

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

Square 1.

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

Multiply -4 times -1.

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

Multiply 4 times \frac{1}{2}.

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

Add 1 to 2.

$$x=\frac{-1±\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±\sqrt{3}}{-2}$$

Now solve the equation x=\frac{-1±\sqrt{3}}{-2} when ± is plus. Add -1 to \sqrt{3}.

$$x=\frac{\sqrt{3}-1}{-2}$$

Divide -1+\sqrt{3} by -2.

$$x=\frac{1-\sqrt{3}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{3}}{-2} when ± is minus. Subtract \sqrt{3} from -1.

$$x=\frac{-\sqrt{3}-1}{-2}$$

Divide -1-\sqrt{3} by -2.

$$x=\frac{\sqrt{3}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{3}}{2} x=\frac{\sqrt{3}+1}{2}$$

Steps for Completing the Square

Add x to both sides.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 1 by -1.

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

Divide -\frac{1}{2} by -1.

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

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.

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

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

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

Add \frac{1}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{3}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{3}}{2} x-\frac{1}{2}=-\frac{\sqrt{3}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{3}+1}{2} x=\frac{1-\sqrt{3}}{2}$$