Solve 0=-1/3x^2+2x+1/2

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

Solve for x

$$x=\frac{\sqrt{42}}{2}+3\approx 6.240370349$$

$$x=-\frac{\sqrt{42}}{2}+3\approx -0.240370349$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

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

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

Square 2.

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

Multiply -4 times -\frac{1}{3}.

$$x=\frac{-2±\sqrt{4+\frac{4}{3}\times \frac{1}{2}}}{2\left(-\frac{1}{3}\right)}$$

Multiply \frac{4}{3} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-2±\sqrt{4+\frac{2}{3}}}{2\left(-\frac{1}{3}\right)}$$

Add 4 to \frac{2}{3}.

$$x=\frac{-2±\sqrt{\frac{14}{3}}}{2\left(-\frac{1}{3}\right)}$$

Take the square root of \frac{14}{3}.

$$x=\frac{-2±\frac{\sqrt{42}}{3}}{2\left(-\frac{1}{3}\right)}$$

Multiply 2 times -\frac{1}{3}.

$$x=\frac{-2±\frac{\sqrt{42}}{3}}{-\frac{2}{3}}$$

Now solve the equation x=\frac{-2±\frac{\sqrt{42}}{3}}{-\frac{2}{3}} when ± is plus. Add -2 to \frac{\sqrt{42}}{3}.

$$x=\frac{\frac{\sqrt{42}}{3}-2}{-\frac{2}{3}}$$

Divide -2+\frac{\sqrt{42}}{3} by -\frac{2}{3} by multiplying -2+\frac{\sqrt{42}}{3} by the reciprocal of -\frac{2}{3}.

$$x=-\frac{\sqrt{42}}{2}+3$$

Now solve the equation x=\frac{-2±\frac{\sqrt{42}}{3}}{-\frac{2}{3}} when ± is minus. Subtract \frac{\sqrt{42}}{3} from -2.

$$x=\frac{-\frac{\sqrt{42}}{3}-2}{-\frac{2}{3}}$$

Divide -2-\frac{\sqrt{42}}{3} by -\frac{2}{3} by multiplying -2-\frac{\sqrt{42}}{3} by the reciprocal of -\frac{2}{3}.

$$x=\frac{\sqrt{42}}{2}+3$$

The equation is now solved.

$$x=-\frac{\sqrt{42}}{2}+3 x=\frac{\sqrt{42}}{2}+3$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Subtract \frac{1}{2} from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by -3.

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

Dividing by -\frac{1}{3} undoes the multiplication by -\frac{1}{3}.

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

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

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

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

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

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -3.

$$x^{2}-6x+9=\frac{3}{2}+9$$

Add \frac{3}{2} to 9.

$$x^{2}-6x+9=\frac{21}{2}$$

Factor x^{2}-6x+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}.

$$\left(x-3\right)^{2}=\frac{21}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{\frac{21}{2}}$$

Simplify.

$$x-3=\frac{\sqrt{42}}{2} x-3=-\frac{\sqrt{42}}{2}$$

Add 3 to both sides of the equation.

$$x=\frac{\sqrt{42}}{2}+3 x=-\frac{\sqrt{42}}{2}+3$$