Solve x=2/3x^2-4/3x

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

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$x=0$$

Steps Using Factoring

Subtract \frac{2}{3}x^{2} from both sides.

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

Add \frac{4}{3}x to both sides.

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

Combine x and \frac{4}{3}x to get \frac{7}{3}x.

$$\frac{7}{3}x-\frac{2}{3}x^{2}=0$$

Factor out x.

$$x\left(\frac{7}{3}-\frac{2}{3}x\right)=0$$

To find equation solutions, solve x=0 and \frac{7-2x}{3}=0.

$$x=0 x=\frac{7}{2}$$

Steps Using the Quadratic Formula

Subtract \frac{2}{3}x^{2} from both sides.

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

Add \frac{4}{3}x to both sides.

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

Combine x and \frac{4}{3}x to get \frac{7}{3}x.

$$\frac{7}{3}x-\frac{2}{3}x^{2}=0$$

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.

$$-\frac{2}{3}x^{2}+\frac{7}{3}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{2}{3} for a, \frac{7}{3} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{7}{3}±\sqrt{\left(\frac{7}{3}\right)^{2}}}{2\left(-\frac{2}{3}\right)}$$

Take the square root of \left(\frac{7}{3}\right)^{2}.

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

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

$$x=\frac{-\frac{7}{3}±\frac{7}{3}}{-\frac{4}{3}}$$

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

$$x=\frac{0}{-\frac{4}{3}}$$

Divide 0 by -\frac{4}{3} by multiplying 0 by the reciprocal of -\frac{4}{3}.

$$x=0$$

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

$$x=-\frac{\frac{14}{3}}{-\frac{4}{3}}$$

Divide -\frac{14}{3} by -\frac{4}{3} by multiplying -\frac{14}{3} by the reciprocal of -\frac{4}{3}.

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

The equation is now solved.

$$x=0 x=\frac{7}{2}$$

Steps for Completing the Square

Subtract \frac{2}{3}x^{2} from both sides.

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

Add \frac{4}{3}x to both sides.

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

Combine x and \frac{4}{3}x to get \frac{7}{3}x.

$$\frac{7}{3}x-\frac{2}{3}x^{2}=0$$

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.

$$-\frac{2}{3}x^{2}+\frac{7}{3}x=0$$

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

$$\frac{-\frac{2}{3}x^{2}+\frac{7}{3}x}{-\frac{2}{3}}=\frac{0}{-\frac{2}{3}}$$

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

$$x^{2}+\frac{\frac{7}{3}}{-\frac{2}{3}}x=\frac{0}{-\frac{2}{3}}$$

Divide \frac{7}{3} by -\frac{2}{3} by multiplying \frac{7}{3} by the reciprocal of -\frac{2}{3}.

$$x^{2}-\frac{7}{2}x=\frac{0}{-\frac{2}{3}}$$

Divide 0 by -\frac{2}{3} by multiplying 0 by the reciprocal of -\frac{2}{3}.

$$x^{2}-\frac{7}{2}x=0$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{7}{4} x-\frac{7}{4}=-\frac{7}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{7}{2} x=0$$