Solve -x+6=-1/3x^2+8/3x

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

Solve for x

$$x=2$$

$$x=9$$

Steps Using the Quadratic Formula

Add \frac{1}{3}x^{2} to both sides.

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

Subtract \frac{8}{3}x from both sides.

$$-x+6+\frac{1}{3}x^{2}-\frac{8}{3}x=0$$

Combine -x and -\frac{8}{3}x to get -\frac{11}{3}x.

$$-\frac{11}{3}x+6+\frac{1}{3}x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{1}{3}x^{2}-\frac{11}{3}x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, -\frac{11}{3} for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{11}{3}\right)±\sqrt{\left(-\frac{11}{3}\right)^{2}-4\times \frac{1}{3}\times 6}}{2\times \frac{1}{3}}$$

Square -\frac{11}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{11}{3}\right)±\sqrt{\frac{121}{9}-4\times \frac{1}{3}\times 6}}{2\times \frac{1}{3}}$$

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

$$x=\frac{-\left(-\frac{11}{3}\right)±\sqrt{\frac{121}{9}-\frac{4}{3}\times 6}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times 6.

$$x=\frac{-\left(-\frac{11}{3}\right)±\sqrt{\frac{121}{9}-8}}{2\times \frac{1}{3}}$$

Add \frac{121}{9} to -8.

$$x=\frac{-\left(-\frac{11}{3}\right)±\sqrt{\frac{49}{9}}}{2\times \frac{1}{3}}$$

Take the square root of \frac{49}{9}.

$$x=\frac{-\left(-\frac{11}{3}\right)±\frac{7}{3}}{2\times \frac{1}{3}}$$

The opposite of -\frac{11}{3} is \frac{11}{3}.

$$x=\frac{\frac{11}{3}±\frac{7}{3}}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x=\frac{\frac{11}{3}±\frac{7}{3}}{\frac{2}{3}}$$

Now solve the equation x=\frac{\frac{11}{3}±\frac{7}{3}}{\frac{2}{3}} when ± is plus. Add \frac{11}{3} to \frac{7}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=9$$

Now solve the equation x=\frac{\frac{11}{3}±\frac{7}{3}}{\frac{2}{3}} when ± is minus. Subtract \frac{7}{3} from \frac{11}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide \frac{4}{3} by \frac{2}{3} by multiplying \frac{4}{3} by the reciprocal of \frac{2}{3}.

$$x=2$$

The equation is now solved.

$$x=9 x=2$$

Steps for Completing the Square

Add \frac{1}{3}x^{2} to both sides.

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

Subtract \frac{8}{3}x from both sides.

$$-x+6+\frac{1}{3}x^{2}-\frac{8}{3}x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

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

Combine -x and -\frac{8}{3}x to get -\frac{11}{3}x.

$$-\frac{11}{3}x+\frac{1}{3}x^{2}=-6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{1}{3}x^{2}-\frac{11}{3}x=-6$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}x^{2}-\frac{11}{3}x}{\frac{1}{3}}=-\frac{6}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$x^{2}+\left(-\frac{\frac{11}{3}}{\frac{1}{3}}\right)x=-\frac{6}{\frac{1}{3}}$$

Divide -\frac{11}{3} by \frac{1}{3} by multiplying -\frac{11}{3} by the reciprocal of \frac{1}{3}.

$$x^{2}-11x=-\frac{6}{\frac{1}{3}}$$

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

$$x^{2}-11x=-18$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-18+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-18+\frac{121}{4}$$

Add -18 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{49}{4}$$

Factor x^{2}-11x+\frac{121}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{7}{2} x-\frac{11}{2}=-\frac{7}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=9 x=2$$