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

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

Solve for x (complex solution)

$$x=\frac{3+\sqrt{39}i}{4}\approx 0.75+1.5612495i$$

$$x=\frac{-\sqrt{39}i+3}{4}\approx 0.75-1.5612495i$$

Steps Using the Quadratic Formula

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-\left(-3\right)±\sqrt{9-48}}{2\times 2}$$

Add 9 to -48.

$$x=\frac{-\left(-3\right)±\sqrt{-39}}{2\times 2}$$

Take the square root of -39.

$$x=\frac{-\left(-3\right)±\sqrt{39}i}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{39}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±\sqrt{39}i}{4}$$

Now solve the equation x=\frac{3±\sqrt{39}i}{4} when ± is plus. Add 3 to i\sqrt{39}.

$$x=\frac{3+\sqrt{39}i}{4}$$

Now solve the equation x=\frac{3±\sqrt{39}i}{4} when ± is minus. Subtract i\sqrt{39} from 3.

$$x=\frac{-\sqrt{39}i+3}{4}$$

The equation is now solved.

$$x=\frac{3+\sqrt{39}i}{4} x=\frac{-\sqrt{39}i+3}{4}$$

Steps for Completing the Square

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

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

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -6 by 2.

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

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.

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

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

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

Add -3 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{39}{16}$$

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}.

$$\left(x-\frac{3}{4}\right)^{2}=-\frac{39}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{-\frac{39}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{39}i}{4} x-\frac{3}{4}=-\frac{\sqrt{39}i}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{3+\sqrt{39}i}{4} x=\frac{-\sqrt{39}i+3}{4}$$