Solve -1/4x^2-1/2x+2=1/3(x-2)

$$- \frac { 1 } { 4 } x ^ { 2 } - \frac { 1 } { 2 } x + 2 = \frac { 1 } { 3 } ( x - 2 )$$

Solve for x

$$x = -\frac{16}{3} = -5\frac{1}{3} \approx -5.333333333$$

$$x=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{3} by x-2.

$$-\frac{1}{4}x^{2}-\frac{1}{2}x+2=\frac{1}{3}x-\frac{2}{3}$$

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

$$-\frac{1}{4}x^{2}-\frac{1}{2}x+2-\frac{1}{3}x=-\frac{2}{3}$$

Combine -\frac{1}{2}x and -\frac{1}{3}x to get -\frac{5}{6}x.

$$-\frac{1}{4}x^{2}-\frac{5}{6}x+2=-\frac{2}{3}$$

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

$$-\frac{1}{4}x^{2}-\frac{5}{6}x+2+\frac{2}{3}=0$$

Add 2 and \frac{2}{3} to get \frac{8}{3}.

$$-\frac{1}{4}x^{2}-\frac{5}{6}x+\frac{8}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{4} for a, -\frac{5}{6} for b, and \frac{8}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{5}{6}\right)±\sqrt{\left(-\frac{5}{6}\right)^{2}-4\left(-\frac{1}{4}\right)\times \frac{8}{3}}}{2\left(-\frac{1}{4}\right)}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{5}{6}\right)±\sqrt{\frac{25}{36}-4\left(-\frac{1}{4}\right)\times \frac{8}{3}}}{2\left(-\frac{1}{4}\right)}$$

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

$$x=\frac{-\left(-\frac{5}{6}\right)±\sqrt{\frac{25}{36}+\frac{8}{3}}}{2\left(-\frac{1}{4}\right)}$$

Add \frac{25}{36} to \frac{8}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{5}{6}\right)±\sqrt{\frac{121}{36}}}{2\left(-\frac{1}{4}\right)}$$

Take the square root of \frac{121}{36}.

$$x=\frac{-\left(-\frac{5}{6}\right)±\frac{11}{6}}{2\left(-\frac{1}{4}\right)}$$

The opposite of -\frac{5}{6} is \frac{5}{6}.

$$x=\frac{\frac{5}{6}±\frac{11}{6}}{2\left(-\frac{1}{4}\right)}$$

Multiply 2 times -\frac{1}{4}.

$$x=\frac{\frac{5}{6}±\frac{11}{6}}{-\frac{1}{2}}$$

Now solve the equation x=\frac{\frac{5}{6}±\frac{11}{6}}{-\frac{1}{2}} when ± is plus. Add \frac{5}{6} to \frac{11}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide \frac{8}{3} by -\frac{1}{2} by multiplying \frac{8}{3} by the reciprocal of -\frac{1}{2}.

$$x=-\frac{16}{3}$$

Now solve the equation x=\frac{\frac{5}{6}±\frac{11}{6}}{-\frac{1}{2}} when ± is minus. Subtract \frac{11}{6} from \frac{5}{6} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=2$$

The equation is now solved.

$$x=-\frac{16}{3} x=2$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{3} by x-2.

$$-\frac{1}{4}x^{2}-\frac{1}{2}x+2=\frac{1}{3}x-\frac{2}{3}$$

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

$$-\frac{1}{4}x^{2}-\frac{1}{2}x+2-\frac{1}{3}x=-\frac{2}{3}$$

Combine -\frac{1}{2}x and -\frac{1}{3}x to get -\frac{5}{6}x.

$$-\frac{1}{4}x^{2}-\frac{5}{6}x+2=-\frac{2}{3}$$

Subtract 2 from both sides.

$$-\frac{1}{4}x^{2}-\frac{5}{6}x=-\frac{2}{3}-2$$

Subtract 2 from -\frac{2}{3} to get -\frac{8}{3}.

$$-\frac{1}{4}x^{2}-\frac{5}{6}x=-\frac{8}{3}$$

Multiply both sides by -4.

$$\frac{-\frac{1}{4}x^{2}-\frac{5}{6}x}{-\frac{1}{4}}=-\frac{\frac{8}{3}}{-\frac{1}{4}}$$

Dividing by -\frac{1}{4} undoes the multiplication by -\frac{1}{4}.

$$x^{2}+\left(-\frac{\frac{5}{6}}{-\frac{1}{4}}\right)x=-\frac{\frac{8}{3}}{-\frac{1}{4}}$$

Divide -\frac{5}{6} by -\frac{1}{4} by multiplying -\frac{5}{6} by the reciprocal of -\frac{1}{4}.

$$x^{2}+\frac{10}{3}x=-\frac{\frac{8}{3}}{-\frac{1}{4}}$$

Divide -\frac{8}{3} by -\frac{1}{4} by multiplying -\frac{8}{3} by the reciprocal of -\frac{1}{4}.

$$x^{2}+\frac{10}{3}x=\frac{32}{3}$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{32}{3}+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{32}{3}+\frac{25}{9}$$

Add \frac{32}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{121}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{121}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{121}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{11}{3} x+\frac{5}{3}=-\frac{11}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$x=2 x=-\frac{16}{3}$$