Solve 5/12x+2=-2/3x^2+1/3x+2

$$\frac{ 5 }{ 12 } x+2 = - \frac{ 2 }{ 3 } { x }^{ 2 } + \frac{ 1 }{ 3 } x+2$$

Solve for x

$$x=-\frac{1}{8}=-0.125$$

$$x=0$$

Steps Using Factoring

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

$$\frac{5}{12}x+2+\frac{2}{3}x^{2}=\frac{1}{3}x+2$$

Subtract \frac{1}{3}x from both sides.

$$\frac{5}{12}x+2+\frac{2}{3}x^{2}-\frac{1}{3}x=2$$

Combine \frac{5}{12}x and -\frac{1}{3}x to get \frac{1}{12}x.

$$\frac{1}{12}x+2+\frac{2}{3}x^{2}=2$$

Subtract 2 from both sides.

$$\frac{1}{12}x+2+\frac{2}{3}x^{2}-2=0$$

Subtract 2 from 2 to get 0.

$$\frac{1}{12}x+\frac{2}{3}x^{2}=0$$

Factor out x.

$$x\left(\frac{1}{12}+\frac{2}{3}x\right)=0$$

To find equation solutions, solve x=0 and \frac{1}{12}+\frac{2x}{3}=0.

$$x=0 x=-\frac{1}{8}$$

Steps Using the Quadratic Formula

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

$$\frac{5}{12}x+2+\frac{2}{3}x^{2}=\frac{1}{3}x+2$$

Subtract \frac{1}{3}x from both sides.

$$\frac{5}{12}x+2+\frac{2}{3}x^{2}-\frac{1}{3}x=2$$

Combine \frac{5}{12}x and -\frac{1}{3}x to get \frac{1}{12}x.

$$\frac{1}{12}x+2+\frac{2}{3}x^{2}=2$$

Subtract 2 from both sides.

$$\frac{1}{12}x+2+\frac{2}{3}x^{2}-2=0$$

Subtract 2 from 2 to get 0.

$$\frac{1}{12}x+\frac{2}{3}x^{2}=0$$

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.

$$\frac{2}{3}x^{2}+\frac{1}{12}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{3} for a, \frac{1}{12} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{12}±\sqrt{\left(\frac{1}{12}\right)^{2}}}{2\times \frac{2}{3}}$$

Take the square root of \left(\frac{1}{12}\right)^{2}.

$$x=\frac{-\frac{1}{12}±\frac{1}{12}}{2\times \frac{2}{3}}$$

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

$$x=\frac{-\frac{1}{12}±\frac{1}{12}}{\frac{4}{3}}$$

Now solve the equation x=\frac{-\frac{1}{12}±\frac{1}{12}}{\frac{4}{3}} when ± is plus. Add -\frac{1}{12} to \frac{1}{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{1}{12}±\frac{1}{12}}{\frac{4}{3}} when ± is minus. Subtract \frac{1}{12} from -\frac{1}{12} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{1}{6}}{\frac{4}{3}}$$

Divide -\frac{1}{6} by \frac{4}{3} by multiplying -\frac{1}{6} by the reciprocal of \frac{4}{3}.

$$x=-\frac{1}{8}$$

The equation is now solved.

$$x=0 x=-\frac{1}{8}$$

Steps for Completing the Square

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

$$\frac{5}{12}x+2+\frac{2}{3}x^{2}=\frac{1}{3}x+2$$

Subtract \frac{1}{3}x from both sides.

$$\frac{5}{12}x+2+\frac{2}{3}x^{2}-\frac{1}{3}x=2$$

Combine \frac{5}{12}x and -\frac{1}{3}x to get \frac{1}{12}x.

$$\frac{1}{12}x+2+\frac{2}{3}x^{2}=2$$

Subtract 2 from both sides.

$$\frac{1}{12}x+\frac{2}{3}x^{2}=2-2$$

Subtract 2 from 2 to get 0.

$$\frac{1}{12}x+\frac{2}{3}x^{2}=0$$

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.

$$\frac{2}{3}x^{2}+\frac{1}{12}x=0$$

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

$$\frac{\frac{2}{3}x^{2}+\frac{1}{12}x}{\frac{2}{3}}=\frac{0}{\frac{2}{3}}$$

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

$$x^{2}+\frac{\frac{1}{12}}{\frac{2}{3}}x=\frac{0}{\frac{2}{3}}$$

Divide \frac{1}{12} by \frac{2}{3} by multiplying \frac{1}{12} by the reciprocal of \frac{2}{3}.

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

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

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

Divide \frac{1}{8}, the coefficient of the x term, by 2 to get \frac{1}{16}. Then add the square of \frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{8}x+\left(\frac{1}{16}\right)^{2}=\left(\frac{1}{16}\right)^{2}$$

Square \frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{1}{256}$$

Factor x^{2}+\frac{1}{8}x+\frac{1}{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}.

$$\left(x+\frac{1}{16}\right)^{2}=\frac{1}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{16}\right)^{2}}=\sqrt{\frac{1}{256}}$$

Simplify.

$$x+\frac{1}{16}=\frac{1}{16} x+\frac{1}{16}=-\frac{1}{16}$$

Subtract \frac{1}{16} from both sides of the equation.

$$x=0 x=-\frac{1}{8}$$