Solve 3(x^2+2x+1/3)=0

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

Solve for x

$$x=\frac{\sqrt{6}}{3}-1\approx -0.183503419$$

$$x=-\frac{\sqrt{6}}{3}-1\approx -1.816496581$$

Steps Using the Quadratic Formula

Divide both sides by 3. Zero divided by any non-zero number gives zero.

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

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

$$x=\frac{-2±\sqrt{2^{2}-4\times \frac{1}{3}}}{2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times \frac{1}{3}}}{2}$$

Multiply -4 times \frac{1}{3}.

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

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

$$x=\frac{-2±\sqrt{\frac{8}{3}}}{2}$$

Take the square root of \frac{8}{3}.

$$x=\frac{-2±\frac{2\sqrt{6}}{3}}{2}$$

Now solve the equation x=\frac{-2±\frac{2\sqrt{6}}{3}}{2} when ± is plus. Add -2 to \frac{2\sqrt{6}}{3}.

$$x=\frac{\frac{2\sqrt{6}}{3}-2}{2}$$

Divide -2+\frac{2\sqrt{6}}{3} by 2.

$$x=\frac{\sqrt{6}}{3}-1$$

Now solve the equation x=\frac{-2±\frac{2\sqrt{6}}{3}}{2} when ± is minus. Subtract \frac{2\sqrt{6}}{3} from -2.

$$x=\frac{-\frac{2\sqrt{6}}{3}-2}{2}$$

Divide -2-\frac{2\sqrt{6}}{3} by 2.

$$x=-\frac{\sqrt{6}}{3}-1$$

The equation is now solved.

$$x=\frac{\sqrt{6}}{3}-1 x=-\frac{\sqrt{6}}{3}-1$$

Steps for Completing the Square

Divide both sides by 3. Zero divided by any non-zero number gives zero.

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

Subtract \frac{1}{3} from both sides. Anything subtracted from zero gives its negation.

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

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

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

Square 1.

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

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

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=\frac{\sqrt{6}}{3} x+1=-\frac{\sqrt{6}}{3}$$

Subtract 1 from both sides of the equation.

$$x=\frac{\sqrt{6}}{3}-1 x=-\frac{\sqrt{6}}{3}-1$$