Solve -2=-0.6x-x^2

$$- 2 = - 0.6 x - x ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{209} - 3}{10} \approx 1.145683229$$

$$x=\frac{-\sqrt{209}-3}{10}\approx -1.745683229$$

Steps Using the Quadratic Formula

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

$$-0.6x-x^{2}=-2$$

Add 2 to both sides.

$$-0.6x-x^{2}+2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-0.6x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -0.6 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-0.6\right)±\sqrt{\left(-0.6\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square -0.6 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-0.6\right)±\sqrt{0.36-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-0.6\right)±\sqrt{0.36+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\left(-0.6\right)±\sqrt{0.36+8}}{2\left(-1\right)}$$

Add 0.36 to 8.

$$x=\frac{-\left(-0.6\right)±\sqrt{8.36}}{2\left(-1\right)}$$

Take the square root of 8.36.

$$x=\frac{-\left(-0.6\right)±\frac{\sqrt{209}}{5}}{2\left(-1\right)}$$

The opposite of -0.6 is 0.6.

$$x=\frac{0.6±\frac{\sqrt{209}}{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0.6±\frac{\sqrt{209}}{5}}{-2}$$

Now solve the equation x=\frac{0.6±\frac{\sqrt{209}}{5}}{-2} when ± is plus. Add 0.6 to \frac{\sqrt{209}}{5}.

$$x=\frac{\sqrt{209}+3}{-2\times 5}$$

Divide \frac{3+\sqrt{209}}{5} by -2.

$$x=\frac{-\sqrt{209}-3}{10}$$

Now solve the equation x=\frac{0.6±\frac{\sqrt{209}}{5}}{-2} when ± is minus. Subtract \frac{\sqrt{209}}{5} from 0.6.

$$x=\frac{3-\sqrt{209}}{-2\times 5}$$

Divide \frac{3-\sqrt{209}}{5} by -2.

$$x=\frac{\sqrt{209}-3}{10}$$

The equation is now solved.

$$x=\frac{-\sqrt{209}-3}{10} x=\frac{\sqrt{209}-3}{10}$$

Steps for Completing the Square

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

$$-0.6x-x^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-0.6x=-2$$

Divide both sides by -1.

$$\frac{-x^{2}-0.6x}{-1}=-\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{0.6}{-1}\right)x=-\frac{2}{-1}$$

Divide -0.6 by -1.

$$x^{2}+0.6x=-\frac{2}{-1}$$

Divide -2 by -1.

$$x^{2}+0.6x=2$$

Divide 0.6, the coefficient of the x term, by 2 to get 0.3. Then add the square of 0.3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+0.6x+0.3^{2}=2+0.3^{2}$$

Square 0.3 by squaring both the numerator and the denominator of the fraction.

$$x^{2}+0.6x+0.09=2+0.09$$

Add 2 to 0.09.

$$x^{2}+0.6x+0.09=2.09$$

Factor x^{2}+0.6x+0.09. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+0.3\right)^{2}=2.09$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+0.3\right)^{2}}=\sqrt{2.09}$$

Simplify.

$$x+0.3=\frac{\sqrt{209}}{10} x+0.3=-\frac{\sqrt{209}}{10}$$

Subtract 0.3 from both sides of the equation.

$$x=\frac{\sqrt{209}-3}{10} x=\frac{-\sqrt{209}-3}{10}$$