Solve 3x^2+4(x+1/3)=0

$$3 x ^ { 2 } + 4 ( x + \frac { 1 } { 3 } ) = 0$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by x+\frac{1}{3}.

$$3x^{2}+4x+\frac{4}{3}=0$$

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 3\times \frac{4}{3}}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-4±\sqrt{16-12\times \frac{4}{3}}}{2\times 3}$$

Multiply -12 times \frac{4}{3}.

$$x=\frac{-4±\sqrt{16-16}}{2\times 3}$$

Add 16 to -16.

$$x=\frac{-4±\sqrt{0}}{2\times 3}$$

Take the square root of 0.

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

Multiply 2 times 3.

$$x=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

Steps for Completing the Square

Use the distributive property to multiply 4 by x+\frac{1}{3}.

$$3x^{2}+4x+\frac{4}{3}=0$$

Subtract \frac{4}{3} from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -\frac{4}{3} by 3.

$$x^{2}+\frac{4}{3}x=-\frac{4}{9}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=-\frac{4}{9}+\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{-4+4}{9}$$

Add -\frac{4}{9} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=0$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{2}{3} from both sides of the equation.

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

The equation is now solved. Solutions are the same.

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