Solve 4x^2/3+2x/9-1/6=0

$$\frac { 4 x ^ { 2 } } { 3 } + \frac { 2 x } { 9 } - \frac { 1 } { 6 } = 0$$

Solve for x

$$x=\frac{\sqrt{19}-1}{12}\approx 0.279908245$$

$$x=\frac{-\sqrt{19}-1}{12}\approx -0.446574912$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 18, the least common multiple of 3,9,6.

$$6\times 4x^{2}+2\times 2x-3=0$$

Multiply 6 and 4 to get 24.

$$24x^{2}+2\times 2x-3=0$$

Multiply 2 and 2 to get 4.

$$24x^{2}+4x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 24 for a, 4 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 24\left(-3\right)}}{2\times 24}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 24\left(-3\right)}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-4±\sqrt{16-96\left(-3\right)}}{2\times 24}$$

Multiply -96 times -3.

$$x=\frac{-4±\sqrt{16+288}}{2\times 24}$$

Add 16 to 288.

$$x=\frac{-4±\sqrt{304}}{2\times 24}$$

Take the square root of 304.

$$x=\frac{-4±4\sqrt{19}}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{-4±4\sqrt{19}}{48}$$

Now solve the equation x=\frac{-4±4\sqrt{19}}{48} when ± is plus. Add -4 to 4\sqrt{19}.

$$x=\frac{4\sqrt{19}-4}{48}$$

Divide -4+4\sqrt{19} by 48.

$$x=\frac{\sqrt{19}-1}{12}$$

Now solve the equation x=\frac{-4±4\sqrt{19}}{48} when ± is minus. Subtract 4\sqrt{19} from -4.

$$x=\frac{-4\sqrt{19}-4}{48}$$

Divide -4-4\sqrt{19} by 48.

$$x=\frac{-\sqrt{19}-1}{12}$$

The equation is now solved.

$$x=\frac{\sqrt{19}-1}{12} x=\frac{-\sqrt{19}-1}{12}$$

Steps for Completing the Square

Multiply both sides of the equation by 18, the least common multiple of 3,9,6.

$$6\times 4x^{2}+2\times 2x-3=0$$

Multiply 6 and 4 to get 24.

$$24x^{2}+2\times 2x-3=0$$

Multiply 2 and 2 to get 4.

$$24x^{2}+4x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$24x^{2}+4x=3$$

Divide both sides by 24.

$$\frac{24x^{2}+4x}{24}=\frac{3}{24}$$

Dividing by 24 undoes the multiplication by 24.

$$x^{2}+\frac{4}{24}x=\frac{3}{24}$$

Reduce the fraction \frac{4}{24} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{1}{6}x=\frac{3}{24}$$

Reduce the fraction \frac{3}{24} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{6}x=\frac{1}{8}$$

Divide \frac{1}{6}, the coefficient of the x term, by 2 to get \frac{1}{12}. Then add the square of \frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{6}x+\left(\frac{1}{12}\right)^{2}=\frac{1}{8}+\left(\frac{1}{12}\right)^{2}$$

Square \frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=\frac{1}{8}+\frac{1}{144}$$

Add \frac{1}{8} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=\frac{19}{144}$$

Factor x^{2}+\frac{1}{6}x+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{12}\right)^{2}=\frac{19}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{12}\right)^{2}}=\sqrt{\frac{19}{144}}$$

Simplify.

$$x+\frac{1}{12}=\frac{\sqrt{19}}{12} x+\frac{1}{12}=-\frac{\sqrt{19}}{12}$$

Subtract \frac{1}{12} from both sides of the equation.

$$x=\frac{\sqrt{19}-1}{12} x=\frac{-\sqrt{19}-1}{12}$$