Solve -x^2-5x=-1/2x+2

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=-4$$

Steps Using the Quadratic Formula

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

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

Combine -5x and \frac{1}{2}x to get -\frac{9}{2}x.

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

Subtract 2 from both sides.

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

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

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

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply 4 times -2.

$$x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{81}{4}-8}}{2\left(-1\right)}$$

Add \frac{81}{4} to -8.

$$x=\frac{-\left(-\frac{9}{2}\right)±\sqrt{\frac{49}{4}}}{2\left(-1\right)}$$

Take the square root of \frac{49}{4}.

$$x=\frac{-\left(-\frac{9}{2}\right)±\frac{7}{2}}{2\left(-1\right)}$$

The opposite of -\frac{9}{2} is \frac{9}{2}.

$$x=\frac{\frac{9}{2}±\frac{7}{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{\frac{9}{2}±\frac{7}{2}}{-2}$$

Now solve the equation x=\frac{\frac{9}{2}±\frac{7}{2}}{-2} when ± is plus. Add \frac{9}{2} to \frac{7}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{\frac{9}{2}±\frac{7}{2}}{-2} when ± is minus. Subtract \frac{7}{2} from \frac{9}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 1 by -2.

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

The equation is now solved.

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

Steps for Completing the Square

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

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

Combine -5x and \frac{1}{2}x to get -\frac{9}{2}x.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -\frac{9}{2} by -1.

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

Divide 2 by -1.

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

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{2}x+\left(\frac{9}{4}\right)^{2}=-2+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=-2+\frac{81}{16}$$

Add -2 to \frac{81}{16}.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{49}{16}$$

Factor x^{2}+\frac{9}{2}x+\frac{81}{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}.

$$\left(x+\frac{9}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x+\frac{9}{4}=\frac{7}{4} x+\frac{9}{4}=-\frac{7}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

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