Solve 9x^2=-3x

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

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=0$$

Steps Using Factoring

Add 3x to both sides.

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

Factor out x.

$$x\left(9x+3\right)=0$$

To find equation solutions, solve x=0 and 9x+3=0.

$$x=0 x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Add 3x to both sides.

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

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

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

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-3±3}{18}$$

Now solve the equation x=\frac{-3±3}{18} when ± is plus. Add -3 to 3.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

Now solve the equation x=\frac{-3±3}{18} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{18}$$

Reduce the fraction \frac{-6}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{3}$$

The equation is now solved.

$$x=0 x=-\frac{1}{3}$$

Steps for Completing the Square

Add 3x to both sides.

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

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

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

Reduce the fraction \frac{3}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{3}x=\frac{0}{9}$$

Divide 0 by 9.

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

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{1}{6} x+\frac{1}{6}=-\frac{1}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=0 x=-\frac{1}{3}$$