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

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

Solve for x

$$x=-3$$

$$x=2$$

Steps Using the Quadratic Formula

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

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

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

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

Combine -2x and \frac{4}{3}x to get -\frac{2}{3}x.

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

Add 2 to both sides.

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

Add 2 and 2 to get 4.

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

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.

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

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

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

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

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

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

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

Multiply \frac{8}{3} times 4.

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{4}{9}+\frac{32}{3}}}{2\left(-\frac{2}{3}\right)}$$

Add \frac{4}{9} to \frac{32}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{2}{3}\right)±\sqrt{\frac{100}{9}}}{2\left(-\frac{2}{3}\right)}$$

Take the square root of \frac{100}{9}.

$$x=\frac{-\left(-\frac{2}{3}\right)±\frac{10}{3}}{2\left(-\frac{2}{3}\right)}$$

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

$$x=\frac{\frac{2}{3}±\frac{10}{3}}{2\left(-\frac{2}{3}\right)}$$

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

$$x=\frac{\frac{2}{3}±\frac{10}{3}}{-\frac{4}{3}}$$

Now solve the equation x=\frac{\frac{2}{3}±\frac{10}{3}}{-\frac{4}{3}} when ± is plus. Add \frac{2}{3} to \frac{10}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=-3$$

Now solve the equation x=\frac{\frac{2}{3}±\frac{10}{3}}{-\frac{4}{3}} when ± is minus. Subtract \frac{10}{3} from \frac{2}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{8}{3}}{-\frac{4}{3}}$$

Divide -\frac{8}{3} by -\frac{4}{3} by multiplying -\frac{8}{3} by the reciprocal of -\frac{4}{3}.

$$x=2$$

The equation is now solved.

$$x=-3 x=2$$

Steps for Completing the Square

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

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

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

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

Combine -2x and \frac{4}{3}x to get -\frac{2}{3}x.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

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

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.

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

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

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

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

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

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

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

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

$$x^{2}+x=6$$

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{25}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{5}{2} x+\frac{1}{2}=-\frac{5}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=2 x=-3$$