Solve -5/12x-2=2/3x^2-4/3x-2

$$- \frac { 5 } { 12 } x - 2 = \frac { 2 } { 3 } x ^ { 2 } - \frac { 4 } { 3 } x - 2$$

Solve for x

$$x = \frac{11}{8} = 1\frac{3}{8} = 1.375$$

$$x=0$$

Steps Using Factoring

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

$$-\frac{5}{12}x-2-\frac{2}{3}x^{2}=-\frac{4}{3}x-2$$

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

$$-\frac{5}{12}x-2-\frac{2}{3}x^{2}+\frac{4}{3}x=-2$$

Combine -\frac{5}{12}x and \frac{4}{3}x to get \frac{11}{12}x.

$$\frac{11}{12}x-2-\frac{2}{3}x^{2}=-2$$

Add 2 to both sides.

$$\frac{11}{12}x-2-\frac{2}{3}x^{2}+2=0$$

Add -2 and 2 to get 0.

$$\frac{11}{12}x-\frac{2}{3}x^{2}=0$$

Factor out x.

$$x\left(\frac{11}{12}-\frac{2}{3}x\right)=0$$

To find equation solutions, solve x=0 and \frac{11}{12}-\frac{2x}{3}=0.

$$x=0 x=\frac{11}{8}$$

Steps Using the Quadratic Formula

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

$$-\frac{5}{12}x-2-\frac{2}{3}x^{2}=-\frac{4}{3}x-2$$

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

$$-\frac{5}{12}x-2-\frac{2}{3}x^{2}+\frac{4}{3}x=-2$$

Combine -\frac{5}{12}x and \frac{4}{3}x to get \frac{11}{12}x.

$$\frac{11}{12}x-2-\frac{2}{3}x^{2}=-2$$

Add 2 to both sides.

$$\frac{11}{12}x-2-\frac{2}{3}x^{2}+2=0$$

Add -2 and 2 to get 0.

$$\frac{11}{12}x-\frac{2}{3}x^{2}=0$$

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.

$$-\frac{2}{3}x^{2}+\frac{11}{12}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{2}{3} for a, \frac{11}{12} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{11}{12}±\sqrt{\left(\frac{11}{12}\right)^{2}}}{2\left(-\frac{2}{3}\right)}$$

Take the square root of \left(\frac{11}{12}\right)^{2}.

$$x=\frac{-\frac{11}{12}±\frac{11}{12}}{2\left(-\frac{2}{3}\right)}$$

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

$$x=\frac{-\frac{11}{12}±\frac{11}{12}}{-\frac{4}{3}}$$

Now solve the equation x=\frac{-\frac{11}{12}±\frac{11}{12}}{-\frac{4}{3}} when ± is plus. Add -\frac{11}{12} to \frac{11}{12} 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{11}{12}±\frac{11}{12}}{-\frac{4}{3}} when ± is minus. Subtract \frac{11}{12} from -\frac{11}{12} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{11}{6}}{-\frac{4}{3}}$$

Divide -\frac{11}{6} by -\frac{4}{3} by multiplying -\frac{11}{6} by the reciprocal of -\frac{4}{3}.

$$x=\frac{11}{8}$$

The equation is now solved.

$$x=0 x=\frac{11}{8}$$

Steps for Completing the Square

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

$$-\frac{5}{12}x-2-\frac{2}{3}x^{2}=-\frac{4}{3}x-2$$

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

$$-\frac{5}{12}x-2-\frac{2}{3}x^{2}+\frac{4}{3}x=-2$$

Combine -\frac{5}{12}x and \frac{4}{3}x to get \frac{11}{12}x.

$$\frac{11}{12}x-2-\frac{2}{3}x^{2}=-2$$

Add 2 to both sides.

$$\frac{11}{12}x-\frac{2}{3}x^{2}=-2+2$$

Add -2 and 2 to get 0.

$$\frac{11}{12}x-\frac{2}{3}x^{2}=0$$

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.

$$-\frac{2}{3}x^{2}+\frac{11}{12}x=0$$

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

$$\frac{-\frac{2}{3}x^{2}+\frac{11}{12}x}{-\frac{2}{3}}=\frac{0}{-\frac{2}{3}}$$

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

$$x^{2}+\frac{\frac{11}{12}}{-\frac{2}{3}}x=\frac{0}{-\frac{2}{3}}$$

Divide \frac{11}{12} by -\frac{2}{3} by multiplying \frac{11}{12} by the reciprocal of -\frac{2}{3}.

$$x^{2}-\frac{11}{8}x=\frac{0}{-\frac{2}{3}}$$

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

$$x^{2}-\frac{11}{8}x=0$$

Divide -\frac{11}{8}, the coefficient of the x term, by 2 to get -\frac{11}{16}. Then add the square of -\frac{11}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{8}x+\left(-\frac{11}{16}\right)^{2}=\left(-\frac{11}{16}\right)^{2}$$

Square -\frac{11}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{8}x+\frac{121}{256}=\frac{121}{256}$$

Factor x^{2}-\frac{11}{8}x+\frac{121}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{16}\right)^{2}=\frac{121}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{16}\right)^{2}}=\sqrt{\frac{121}{256}}$$

Simplify.

$$x-\frac{11}{16}=\frac{11}{16} x-\frac{11}{16}=-\frac{11}{16}$$

Add \frac{11}{16} to both sides of the equation.

$$x=\frac{11}{8} x=0$$