Solve left(3xright)^2+2x+9=0

$${ \left(3x \right) }^{ 2 } +2x+9=0$$

Solve for x (complex solution)

$$x=\frac{-1+4\sqrt{5}i}{9}\approx -0.111111111+0.99380799i$$

$$x=\frac{-4\sqrt{5}i-1}{9}\approx -0.111111111-0.99380799i$$

Steps Using the Quadratic Formula

Expand \left(3x\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

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

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

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

Square 2.

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

Multiply -4 times 9.

$$x=\frac{-2±\sqrt{4-36\times 9}}{2\times 9}$$

Multiply -36 times 9.

$$x=\frac{-2±\sqrt{4-324}}{2\times 9}$$

Add 4 to -324.

$$x=\frac{-2±\sqrt{-320}}{2\times 9}$$

Take the square root of -320.

$$x=\frac{-2±8\sqrt{5}i}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-2±8\sqrt{5}i}{18}$$

Now solve the equation x=\frac{-2±8\sqrt{5}i}{18} when ± is plus. Add -2 to 8i\sqrt{5}.

$$x=\frac{-2+8\sqrt{5}i}{18}$$

Divide -2+8i\sqrt{5} by 18.

$$x=\frac{-1+4\sqrt{5}i}{9}$$

Now solve the equation x=\frac{-2±8\sqrt{5}i}{18} when ± is minus. Subtract 8i\sqrt{5} from -2.

$$x=\frac{-8\sqrt{5}i-2}{18}$$

Divide -2-8i\sqrt{5} by 18.

$$x=\frac{-4\sqrt{5}i-1}{9}$$

The equation is now solved.

$$x=\frac{-1+4\sqrt{5}i}{9} x=\frac{-4\sqrt{5}i-1}{9}$$

Steps for Completing the Square

Expand \left(3x\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

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

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

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

Divide -9 by 9.

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

Divide \frac{2}{9}, the coefficient of the x term, by 2 to get \frac{1}{9}. Then add the square of \frac{1}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{9}x+\frac{1}{81}=-1+\frac{1}{81}$$

Add -1 to \frac{1}{81}.

$$x^{2}+\frac{2}{9}x+\frac{1}{81}=-\frac{80}{81}$$

Factor x^{2}+\frac{2}{9}x+\frac{1}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{9}\right)^{2}=-\frac{80}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{9}\right)^{2}}=\sqrt{-\frac{80}{81}}$$

Simplify.

$$x+\frac{1}{9}=\frac{4\sqrt{5}i}{9} x+\frac{1}{9}=-\frac{4\sqrt{5}i}{9}$$

Subtract \frac{1}{9} from both sides of the equation.

$$x=\frac{-1+4\sqrt{5}i}{9} x=\frac{-4\sqrt{5}i-1}{9}$$