Solve -2/9=x^2-x

$$- \frac { 2 } { 9 } = x ^ { 2 } - x$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using the Quadratic Formula

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

$$x^{2}-x=-\frac{2}{9}$$

Add \frac{2}{9} to both sides.

$$x^{2}-x+\frac{2}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and \frac{2}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{2}{9}}}{2}$$

Multiply -4 times \frac{2}{9}.

$$x=\frac{-\left(-1\right)±\sqrt{1-\frac{8}{9}}}{2}$$

Add 1 to -\frac{8}{9}.

$$x=\frac{-\left(-1\right)±\sqrt{\frac{1}{9}}}{2}$$

Take the square root of \frac{1}{9}.

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

The opposite of -1 is 1.

$$x=\frac{1±\frac{1}{3}}{2}$$

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

$$x=\frac{\frac{4}{3}}{2}$$

Divide \frac{4}{3} by 2.

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

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

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

Divide \frac{2}{3} by 2.

$$x=\frac{1}{3}$$

The equation is now solved.

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

Steps for Completing the Square

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

$$x^{2}-x=-\frac{2}{9}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{2}{9}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=-\frac{2}{9}+\frac{1}{4}$$

Add -\frac{2}{9} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{1}{36}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{1}{2}=\frac{1}{6} x-\frac{1}{2}=-\frac{1}{6}$$

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

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