Solve (2/3x^2-1/6x-1/2)*6=0

$$( \frac{ 2 }{ 3 } { x }^{ 2 } - \frac{ 1 }{ 6 } x- \frac{ 1 }{ 2 } ) \times 6 = 0$$

Solve for x

$$x=-\frac{3}{4}=-0.75$$

$$x=1$$

Steps Using the Quadratic Formula

Divide both sides by 6. Zero divided by any non-zero number gives zero.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{3} for a, -\frac{1}{6} for b, and -\frac{1}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{1}{6}\right)±\sqrt{\left(-\frac{1}{6}\right)^{2}-4\times \frac{2}{3}\left(-\frac{1}{2}\right)}}{2\times \frac{2}{3}}$$

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

$$x=\frac{-\left(-\frac{1}{6}\right)±\sqrt{\frac{1}{36}-4\times \frac{2}{3}\left(-\frac{1}{2}\right)}}{2\times \frac{2}{3}}$$

Multiply -4 times \frac{2}{3}.

$$x=\frac{-\left(-\frac{1}{6}\right)±\sqrt{\frac{1}{36}-\frac{8}{3}\left(-\frac{1}{2}\right)}}{2\times \frac{2}{3}}$$

Multiply -\frac{8}{3} times -\frac{1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{1}{6}\right)±\sqrt{\frac{1}{36}+\frac{4}{3}}}{2\times \frac{2}{3}}$$

Add \frac{1}{36} to \frac{4}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{1}{6}\right)±\sqrt{\frac{49}{36}}}{2\times \frac{2}{3}}$$

Take the square root of \frac{49}{36}.

$$x=\frac{-\left(-\frac{1}{6}\right)±\frac{7}{6}}{2\times \frac{2}{3}}$$

The opposite of -\frac{1}{6} is \frac{1}{6}.

$$x=\frac{\frac{1}{6}±\frac{7}{6}}{2\times \frac{2}{3}}$$

Multiply 2 times \frac{2}{3}.

$$x=\frac{\frac{1}{6}±\frac{7}{6}}{\frac{4}{3}}$$

Now solve the equation x=\frac{\frac{1}{6}±\frac{7}{6}}{\frac{4}{3}} when ± is plus. Add \frac{1}{6} to \frac{7}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{4}{3}}{\frac{4}{3}}$$

Divide \frac{4}{3} by \frac{4}{3} by multiplying \frac{4}{3} by the reciprocal of \frac{4}{3}.

$$x=1$$

Now solve the equation x=\frac{\frac{1}{6}±\frac{7}{6}}{\frac{4}{3}} when ± is minus. Subtract \frac{7}{6} from \frac{1}{6} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{1}{\frac{4}{3}}$$

Divide -1 by \frac{4}{3} by multiplying -1 by the reciprocal of \frac{4}{3}.

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

The equation is now solved.

$$x=1 x=-\frac{3}{4}$$

Steps for Completing the Square

Divide both sides by 6. Zero divided by any non-zero number gives zero.

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

Add \frac{1}{2} to both sides. Anything plus zero gives itself.

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

Divide both sides of the equation by \frac{2}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by \frac{2}{3} undoes the multiplication by \frac{2}{3}.

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

Divide -\frac{1}{6} by \frac{2}{3} by multiplying -\frac{1}{6} by the reciprocal of \frac{2}{3}.

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

Divide \frac{1}{2} by \frac{2}{3} by multiplying \frac{1}{2} by the reciprocal of \frac{2}{3}.

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

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{3}{4}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{3}{4}+\frac{1}{64}$$

Add \frac{3}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{49}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{7}{8} x-\frac{1}{8}=-\frac{7}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=1 x=-\frac{3}{4}$$