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

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

Solve for x (complex solution)

$$x=\frac{-3+3\sqrt{15}i}{4}\approx -0.75+2.90473751i$$

$$x=\frac{-3\sqrt{15}i-3}{4}\approx -0.75-2.90473751i$$

Steps Using the Quadratic Formula

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

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

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

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

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

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\times \frac{1}{3}\times 3}}{2\times \frac{1}{3}}$$

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

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-\frac{4}{3}\times 3}}{2\times \frac{1}{3}}$$

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

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4}}{2\times \frac{1}{3}}$$

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

$$x=\frac{-\frac{1}{2}±\sqrt{-\frac{15}{4}}}{2\times \frac{1}{3}}$$

Take the square root of -\frac{15}{4}.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{15}i}{2}}{2\times \frac{1}{3}}$$

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

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{15}i}{2}}{\frac{2}{3}}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{15}i}{2}}{\frac{2}{3}} when ± is plus. Add -\frac{1}{2} to \frac{i\sqrt{15}}{2}.

$$x=\frac{-1+\sqrt{15}i}{\frac{2}{3}\times 2}$$

Divide \frac{-1+i\sqrt{15}}{2} by \frac{2}{3} by multiplying \frac{-1+i\sqrt{15}}{2} by the reciprocal of \frac{2}{3}.

$$x=\frac{-3+3\sqrt{15}i}{4}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{15}i}{2}}{\frac{2}{3}} when ± is minus. Subtract \frac{i\sqrt{15}}{2} from -\frac{1}{2}.

$$x=\frac{-\sqrt{15}i-1}{\frac{2}{3}\times 2}$$

Divide \frac{-1-i\sqrt{15}}{2} by \frac{2}{3} by multiplying \frac{-1-i\sqrt{15}}{2} by the reciprocal of \frac{2}{3}.

$$x=\frac{-3\sqrt{15}i-3}{4}$$

The equation is now solved.

$$x=\frac{-3+3\sqrt{15}i}{4} x=\frac{-3\sqrt{15}i-3}{4}$$

Steps for Completing the Square

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

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by 3.

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

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

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

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

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

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

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

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-9+\frac{9}{16}$$

Add -9 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{135}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{4}\right)^{2}=-\frac{135}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{-\frac{135}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{3\sqrt{15}i}{4} x+\frac{3}{4}=-\frac{3\sqrt{15}i}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{-3+3\sqrt{15}i}{4} x=\frac{-3\sqrt{15}i-3}{4}$$