Solve x=-3x^2

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

Solve for x

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

$$x=0$$

Steps Using Factoring

Add 3x^{2} to both sides.

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

Factor out x.

$$x\left(1+3x\right)=0$$

To find equation solutions, solve x=0 and 1+3x=0.

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

Steps Using the Quadratic Formula

Add 3x^{2} to both sides.

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

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.

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

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

$$x=\frac{-1±\sqrt{1^{2}}}{2\times 3}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±1}{6}$$

Now solve the equation x=\frac{-1±1}{6} when ± is plus. Add -1 to 1.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Now solve the equation x=\frac{-1±1}{6} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{6}$$

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Add 3x^{2} to both sides.

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

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.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 0 by 3.

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

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.

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