Solve 1/6x(3x-6)=8-1/3x

$$\frac{ 1 }{ 6 } x(3x-6)=8- \frac{ 1 }{ 3 } x$$

Solve for x

$$x = \frac{2 \sqrt{37} + 2}{3} \approx 4.721841687$$

$$x=\frac{2-2\sqrt{37}}{3}\approx -3.388508354$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{6}x by 3x-6.

$$\frac{1}{6}x\times 3x+\frac{1}{6}x\left(-6\right)=8-\frac{1}{3}x$$

Multiply x and x to get x^{2}.

$$\frac{1}{6}x^{2}\times 3+\frac{1}{6}x\left(-6\right)=8-\frac{1}{3}x$$

Multiply \frac{1}{6} and 3 to get \frac{3}{6}.

$$\frac{3}{6}x^{2}+\frac{1}{6}x\left(-6\right)=8-\frac{1}{3}x$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$\frac{1}{2}x^{2}+\frac{1}{6}x\left(-6\right)=8-\frac{1}{3}x$$

Multiply \frac{1}{6} and -6 to get \frac{-6}{6}.

$$\frac{1}{2}x^{2}+\frac{-6}{6}x=8-\frac{1}{3}x$$

Divide -6 by 6 to get -1.

$$\frac{1}{2}x^{2}-x=8-\frac{1}{3}x$$

Subtract 8 from both sides.

$$\frac{1}{2}x^{2}-x-8=-\frac{1}{3}x$$

Add \frac{1}{3}x to both sides.

$$\frac{1}{2}x^{2}-x-8+\frac{1}{3}x=0$$

Combine -x and \frac{1}{3}x to get -\frac{2}{3}x.

$$\frac{1}{2}x^{2}-\frac{2}{3}x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -\frac{2}{3} for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}-4\times \frac{1}{2}\left(-8\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}-2\left(-8\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -8.

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}+16}}{2\times \frac{1}{2}}$$

Add \frac{4}{9} to 16.

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{148}{9}}}{2\times \frac{1}{2}}$$

Take the square root of \frac{148}{9}.

$$x=\frac{-\left(-\frac{2}{3}\right)±\frac{2\sqrt{37}}{3}}{2\times \frac{1}{2}}$$

The opposite of -\frac{2}{3} is \frac{2}{3}.

$$x=\frac{\frac{2}{3}±\frac{2\sqrt{37}}{3}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{\frac{2}{3}±\frac{2\sqrt{37}}{3}}{1}$$

Now solve the equation x=\frac{\frac{2}{3}±\frac{2\sqrt{37}}{3}}{1} when ± is plus. Add \frac{2}{3} to \frac{2\sqrt{37}}{3}.

$$x=\frac{2\sqrt{37}+2}{3}$$

Now solve the equation x=\frac{\frac{2}{3}±\frac{2\sqrt{37}}{3}}{1} when ± is minus. Subtract \frac{2\sqrt{37}}{3} from \frac{2}{3}.

$$x=\frac{2-2\sqrt{37}}{3}$$

The equation is now solved.

$$x=\frac{2\sqrt{37}+2}{3} x=\frac{2-2\sqrt{37}}{3}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{6}x by 3x-6.

$$\frac{1}{6}x\times 3x+\frac{1}{6}x\left(-6\right)=8-\frac{1}{3}x$$

Multiply x and x to get x^{2}.

$$\frac{1}{6}x^{2}\times 3+\frac{1}{6}x\left(-6\right)=8-\frac{1}{3}x$$

Multiply \frac{1}{6} and 3 to get \frac{3}{6}.

$$\frac{3}{6}x^{2}+\frac{1}{6}x\left(-6\right)=8-\frac{1}{3}x$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$\frac{1}{2}x^{2}+\frac{1}{6}x\left(-6\right)=8-\frac{1}{3}x$$

Multiply \frac{1}{6} and -6 to get \frac{-6}{6}.

$$\frac{1}{2}x^{2}+\frac{-6}{6}x=8-\frac{1}{3}x$$

Divide -6 by 6 to get -1.

$$\frac{1}{2}x^{2}-x=8-\frac{1}{3}x$$

Add \frac{1}{3}x to both sides.

$$\frac{1}{2}x^{2}-x+\frac{1}{3}x=8$$

Combine -x and \frac{1}{3}x to get -\frac{2}{3}x.

$$\frac{1}{2}x^{2}-\frac{2}{3}x=8$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-\frac{2}{3}x}{\frac{1}{2}}=\frac{8}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

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

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

$$x^{2}-\frac{4}{3}x=\frac{8}{\frac{1}{2}}$$

Divide 8 by \frac{1}{2} by multiplying 8 by the reciprocal of \frac{1}{2}.

$$x^{2}-\frac{4}{3}x=16$$

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.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=16+\left(-\frac{2}{3}\right)^{2}$$

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=16+\frac{4}{9}$$

Add 16 to \frac{4}{9}.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{148}{9}$$

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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{148}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{148}{9}}$$

Simplify.

$$x-\frac{2}{3}=\frac{2\sqrt{37}}{3} x-\frac{2}{3}=-\frac{2\sqrt{37}}{3}$$

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{2\sqrt{37}+2}{3} x=\frac{2-2\sqrt{37}}{3}$$