Solve 0=3x^2-3x+16

$$0 = 3 x ^ { 2 } - 3 x + 16$$

Solve for x (complex solution)

$$x=\frac{\sqrt{183}i}{6}+\frac{1}{2}\approx 0.5+2.254624876i$$

$$x=-\frac{\sqrt{183}i}{6}+\frac{1}{2}\approx 0.5-2.254624876i$$

Steps Using the Quadratic Formula

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

$$3x^{2}-3x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -3 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\times 16}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-3\right)±\sqrt{9-12\times 16}}{2\times 3}$$

Multiply -12 times 16.

$$x=\frac{-\left(-3\right)±\sqrt{9-192}}{2\times 3}$$

Add 9 to -192.

$$x=\frac{-\left(-3\right)±\sqrt{-183}}{2\times 3}$$

Take the square root of -183.

$$x=\frac{-\left(-3\right)±\sqrt{183}i}{2\times 3}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{183}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{3±\sqrt{183}i}{6}$$

Now solve the equation x=\frac{3±\sqrt{183}i}{6} when ± is plus. Add 3 to i\sqrt{183}.

$$x=\frac{3+\sqrt{183}i}{6}$$

Divide 3+i\sqrt{183} by 6.

$$x=\frac{\sqrt{183}i}{6}+\frac{1}{2}$$

Now solve the equation x=\frac{3±\sqrt{183}i}{6} when ± is minus. Subtract i\sqrt{183} from 3.

$$x=\frac{-\sqrt{183}i+3}{6}$$

Divide 3-i\sqrt{183} by 6.

$$x=-\frac{\sqrt{183}i}{6}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{183}i}{6}+\frac{1}{2} x=-\frac{\sqrt{183}i}{6}+\frac{1}{2}$$

Steps for Completing the Square

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

$$3x^{2}-3x+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-3x=-16$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -3 by 3.

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

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{16}{3}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=-\frac{16}{3}+\frac{1}{4}$$

Add -\frac{16}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=-\frac{61}{12}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{61}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{61}{12}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{183}i}{6} x-\frac{1}{2}=-\frac{\sqrt{183}i}{6}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{183}i}{6}+\frac{1}{2} x=-\frac{\sqrt{183}i}{6}+\frac{1}{2}$$