Solve -x^2-3x-2/2=x

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

Solve for x

$$x=\frac{\sqrt{17}-5}{2}\approx -0.438447187$$

$$x=\frac{-\sqrt{17}-5}{2}\approx -4.561552813$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$\frac{-x^{2}-3x-2}{2}-x=0$$

Multiply both sides of the equation by 2.

$$-x^{2}-3x-2-2x=0$$

Reorder the terms.

$$-x^{2}-3x-2x-2=0$$

Combine -3x and -2x to get -5x.

$$-x^{2}-5x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply 4 times -2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8}}{2\left(-1\right)}$$

Add 25 to -8.

$$x=\frac{-\left(-5\right)±\sqrt{17}}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{17}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±\sqrt{17}}{-2}$$

Now solve the equation x=\frac{5±\sqrt{17}}{-2} when ± is plus. Add 5 to \sqrt{17}.

$$x=\frac{\sqrt{17}+5}{-2}$$

Divide 5+\sqrt{17} by -2.

$$x=\frac{-\sqrt{17}-5}{2}$$

Now solve the equation x=\frac{5±\sqrt{17}}{-2} when ± is minus. Subtract \sqrt{17} from 5.

$$x=\frac{5-\sqrt{17}}{-2}$$

Divide 5-\sqrt{17} by -2.

$$x=\frac{\sqrt{17}-5}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{17}-5}{2} x=\frac{\sqrt{17}-5}{2}$$

Steps for Completing the Square

Subtract x from both sides.

$$\frac{-x^{2}-3x-2}{2}-x=0$$

Multiply both sides of the equation by 2.

$$-x^{2}-3x-2-2x=0$$

Reorder the terms.

$$-x^{2}-3x-2x-2=0$$

Combine -3x and -2x to get -5x.

$$-x^{2}-5x-2=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$-x^{2}-5x=2$$

Divide both sides by -1.

$$\frac{-x^{2}-5x}{-1}=\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{5}{-1}\right)x=\frac{2}{-1}$$

Divide -5 by -1.

$$x^{2}+5x=\frac{2}{-1}$$

Divide 2 by -1.

$$x^{2}+5x=-2$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-2+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-2+\frac{25}{4}$$

Add -2 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{17}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{17}}{2} x+\frac{5}{2}=-\frac{\sqrt{17}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{17}-5}{2} x=\frac{-\sqrt{17}-5}{2}$$