Solve x(x+3)-1/2=-2

$$x ( x + 3 ) - 1 / 2 = - 2$$

Solve for x

$$x=\frac{\sqrt{3}-3}{2}\approx -0.633974596$$

$$x=\frac{-\sqrt{3}-3}{2}\approx -2.366025404$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x+3.

$$x^{2}+3x-\frac{1}{2}=-2$$

Add 2 to both sides.

$$x^{2}+3x-\frac{1}{2}+2=0$$

Convert 2 to fraction \frac{4}{2}.

$$x^{2}+3x-\frac{1}{2}+\frac{4}{2}=0$$

Since -\frac{1}{2} and \frac{4}{2} have the same denominator, add them by adding their numerators.

$$x^{2}+3x+\frac{-1+4}{2}=0$$

Add -1 and 4 to get 3.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and \frac{3}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times \frac{3}{2}}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times \frac{3}{2}}}{2}$$

Multiply -4 times \frac{3}{2}.

$$x=\frac{-3±\sqrt{9-6}}{2}$$

Add 9 to -6.

$$x=\frac{-3±\sqrt{3}}{2}$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply x by x+3.

$$x^{2}+3x-\frac{1}{2}=-2$$

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

$$x^{2}+3x=-2+\frac{1}{2}$$

Convert -2 to fraction -\frac{4}{2}.

$$x^{2}+3x=-\frac{4}{2}+\frac{1}{2}$$

Since -\frac{4}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$x^{2}+3x=\frac{-4+1}{2}$$

Add -4 and 1 to get -3.

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

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-\frac{3}{2}+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-\frac{3}{2}+\frac{9}{4}$$

Add -\frac{3}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{3}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{3}{4}}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

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