Solve 3x(x+1)=-9

$$3x(x+1)=-9$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{11}i}{2}\approx -0.5+1.658312395i$$

$$x=\frac{-\sqrt{11}i-1}{2}\approx -0.5-1.658312395i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by x+1.

$$3x^{2}+3x=-9$$

Add 9 to both sides.

$$3x^{2}+3x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 3 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3.

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

Multiply -4 times 3.

$$x=\frac{-3±\sqrt{9-12\times 9}}{2\times 3}$$

Multiply -12 times 9.

$$x=\frac{-3±\sqrt{9-108}}{2\times 3}$$

Add 9 to -108.

$$x=\frac{-3±\sqrt{-99}}{2\times 3}$$

Take the square root of -99.

$$x=\frac{-3±3\sqrt{11}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-3±3\sqrt{11}i}{6}$$

Now solve the equation x=\frac{-3±3\sqrt{11}i}{6} when ± is plus. Add -3 to 3i\sqrt{11}.

$$x=\frac{-3+3\sqrt{11}i}{6}$$

Divide -3+3i\sqrt{11} by 6.

$$x=\frac{-1+\sqrt{11}i}{2}$$

Now solve the equation x=\frac{-3±3\sqrt{11}i}{6} when ± is minus. Subtract 3i\sqrt{11} from -3.

$$x=\frac{-3\sqrt{11}i-3}{6}$$

Divide -3-3i\sqrt{11} by 6.

$$x=\frac{-\sqrt{11}i-1}{2}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 3x by x+1.

$$3x^{2}+3x=-9$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 3 by 3.

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

Divide -9 by 3.

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

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-3+\left(\frac{1}{2}\right)^{2}$$

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

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

Add -3 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=-\frac{11}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{11}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{11}i}{2} x+\frac{1}{2}=-\frac{\sqrt{11}i}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{-1+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-1}{2}$$