Solve 0=-3x^2+x-4?

$$0 = - 3 x ^ { 2 } + x - 4 ?$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{47}i+1}{6}\approx 0.166666667-1.1426091i$$

$$x=\frac{1+\sqrt{47}i}{6}\approx 0.166666667+1.1426091i$$

Steps Using the Quadratic Formula

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

$$-3x^{2}+x-4=0$$

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

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

Square 1.

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

Multiply -4 times -3.

$$x=\frac{-1±\sqrt{1+12\left(-4\right)}}{2\left(-3\right)}$$

Multiply 12 times -4.

$$x=\frac{-1±\sqrt{1-48}}{2\left(-3\right)}$$

Add 1 to -48.

$$x=\frac{-1±\sqrt{-47}}{2\left(-3\right)}$$

Take the square root of -47.

$$x=\frac{-1±\sqrt{47}i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-1±\sqrt{47}i}{-6}$$

Now solve the equation x=\frac{-1±\sqrt{47}i}{-6} when ± is plus. Add -1 to i\sqrt{47}.

$$x=\frac{-1+\sqrt{47}i}{-6}$$

Divide -1+i\sqrt{47} by -6.

$$x=\frac{-\sqrt{47}i+1}{6}$$

Now solve the equation x=\frac{-1±\sqrt{47}i}{-6} when ± is minus. Subtract i\sqrt{47} from -1.

$$x=\frac{-\sqrt{47}i-1}{-6}$$

Divide -1-i\sqrt{47} by -6.

$$x=\frac{1+\sqrt{47}i}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{47}i+1}{6} x=\frac{1+\sqrt{47}i}{6}$$

Steps for Completing the Square

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

$$-3x^{2}+x-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$-3x^{2}+x=4$$

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 1 by -3.

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

Divide 4 by -3.

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

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{4}{3}+\frac{1}{36}$$

Add -\frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{47}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=-\frac{47}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{-\frac{47}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{47}i}{6} x-\frac{1}{6}=-\frac{\sqrt{47}i}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{1+\sqrt{47}i}{6} x=\frac{-\sqrt{47}i+1}{6}$$