Solve 5x^2-2x+3/6x=0

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{14}i}{5}\approx 0.2+0.748331477i$$

$$x=\frac{-\sqrt{14}i+1}{5}\approx 0.2-0.748331477i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x.

$$5x^{2}-2x+3=0$$

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

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

Square -2.

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

Multiply -4 times 5.

$$x=\frac{-\left(-2\right)±\sqrt{4-20\times 3}}{2\times 5}$$

Multiply -20 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-60}}{2\times 5}$$

Add 4 to -60.

$$x=\frac{-\left(-2\right)±\sqrt{-56}}{2\times 5}$$

Take the square root of -56.

$$x=\frac{-\left(-2\right)±2\sqrt{14}i}{2\times 5}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{14}i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{2±2\sqrt{14}i}{10}$$

Now solve the equation x=\frac{2±2\sqrt{14}i}{10} when ± is plus. Add 2 to 2i\sqrt{14}.

$$x=\frac{2+2\sqrt{14}i}{10}$$

Divide 2+2i\sqrt{14} by 10.

$$x=\frac{1+\sqrt{14}i}{5}$$

Now solve the equation x=\frac{2±2\sqrt{14}i}{10} when ± is minus. Subtract 2i\sqrt{14} from 2.

$$x=\frac{-2\sqrt{14}i+2}{10}$$

Divide 2-2i\sqrt{14} by 10.

$$x=\frac{-\sqrt{14}i+1}{5}$$

The equation is now solved.

$$x=\frac{1+\sqrt{14}i}{5} x=\frac{-\sqrt{14}i+1}{5}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x.

$$5x^{2}-2x+3=0$$

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

$$5x^{2}-2x=-3$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{3}{5}+\frac{1}{25}$$

Add -\frac{3}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{14}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=-\frac{14}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{-\frac{14}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{\sqrt{14}i}{5} x-\frac{1}{5}=-\frac{\sqrt{14}i}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{1+\sqrt{14}i}{5} x=\frac{-\sqrt{14}i+1}{5}$$