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

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

Solve for x

$$x = \frac{\sqrt{33} - 1}{4} \approx 1.186140662$$

$$x=\frac{-\sqrt{33}-1}{4}\approx -1.686140662$$

Steps Using the Quadratic Formula

Subtract \frac{1}{2} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of -x and 2 is 2x. Multiply \frac{-x^{2}-x+2}{-x} times \frac{-2}{-2}. Multiply \frac{1}{2} times \frac{x}{x}.

$$\frac{-2\left(-x^{2}-x+2\right)}{2x}-\frac{x}{2x}=0$$

Since \frac{-2\left(-x^{2}-x+2\right)}{2x} and \frac{x}{2x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-2\left(-x^{2}-x+2\right)-x}{2x}=0$$

Do the multiplications in -2\left(-x^{2}-x+2\right)-x.

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

Combine like terms in 2x^{2}+2x-4-x.

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

Factor the expressions that are not already factored in \frac{2x^{2}+x-4}{2x}.

$$\frac{2\left(x-\left(-\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)\left(x-\left(\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)}{2x}=0$$

Cancel out 2 in both numerator and denominator.

$$\frac{\left(x-\left(-\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)\left(x-\left(\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)}{x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$\left(x-\left(-\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)\left(x-\left(\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)=0$$

To find the opposite of -\frac{1}{4}\sqrt{33}-\frac{1}{4}, find the opposite of each term.

$$\left(x+\frac{1}{4}\sqrt{33}+\frac{1}{4}\right)\left(x-\left(\frac{1}{4}\sqrt{33}-\frac{1}{4}\right)\right)=0$$

To find the opposite of \frac{1}{4}\sqrt{33}-\frac{1}{4}, find the opposite of each term.

$$\left(x+\frac{1}{4}\sqrt{33}+\frac{1}{4}\right)\left(x-\frac{1}{4}\sqrt{33}+\frac{1}{4}\right)=0$$

Use the distributive property to multiply x+\frac{1}{4}\sqrt{33}+\frac{1}{4} by x-\frac{1}{4}\sqrt{33}+\frac{1}{4} and combine like terms.

$$x^{2}+\frac{1}{2}x-\frac{1}{16}\left(\sqrt{33}\right)^{2}+\frac{1}{16}=0$$

The square of \sqrt{33} is 33.

$$x^{2}+\frac{1}{2}x-\frac{1}{16}\times 33+\frac{1}{16}=0$$

Multiply -\frac{1}{16} and 33 to get -\frac{33}{16}.

$$x^{2}+\frac{1}{2}x-\frac{33}{16}+\frac{1}{16}=0$$

Add -\frac{33}{16} and \frac{1}{16} to get -2.

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

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

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

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

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

Multiply -4 times -2.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+8}}{2}$$

Add \frac{1}{4} to 8.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{33}{4}}}{2}$$

Take the square root of \frac{33}{4}.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{33}}{2}}{2}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{33}}{2}}{2} when ± is plus. Add -\frac{1}{2} to \frac{\sqrt{33}}{2}.

$$x=\frac{\sqrt{33}-1}{2\times 2}$$

Divide \frac{-1+\sqrt{33}}{2} by 2.

$$x=\frac{\sqrt{33}-1}{4}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{33}}{2}}{2} when ± is minus. Subtract \frac{\sqrt{33}}{2} from -\frac{1}{2}.

$$x=\frac{-\sqrt{33}-1}{2\times 2}$$

Divide \frac{-1-\sqrt{33}}{2} by 2.

$$x=\frac{-\sqrt{33}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{33}-1}{4} x=\frac{-\sqrt{33}-1}{4}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of -x,2.

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

Use the distributive property to multiply -2 by -x^{2}-x+2.

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

Subtract x from both sides.

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

Combine 2x and -x to get x.

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

Add 4 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 4 by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=2+\frac{1}{16}$$

Add 2 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{33}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{33}}{4} x+\frac{1}{4}=-\frac{\sqrt{33}}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{33}-1}{4} x=\frac{-\sqrt{33}-1}{4}$$