Solve 2x^2=5x-8

$$2 x ^ { 2 } = 5 x - 8$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{39}i}{4}\approx 1.25+1.5612495i$$

$$x=\frac{-\sqrt{39}i+5}{4}\approx 1.25-1.5612495i$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

$$2x^{2}-5x=-8$$

Add 8 to both sides.

$$2x^{2}-5x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -5 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\times 8}}{2\times 2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 8}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\times 8}}{2\times 2}$$

Multiply -8 times 8.

$$x=\frac{-\left(-5\right)±\sqrt{25-64}}{2\times 2}$$

Add 25 to -64.

$$x=\frac{-\left(-5\right)±\sqrt{-39}}{2\times 2}$$

Take the square root of -39.

$$x=\frac{-\left(-5\right)±\sqrt{39}i}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{39}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±\sqrt{39}i}{4}$$

Now solve the equation x=\frac{5±\sqrt{39}i}{4} when ± is plus. Add 5 to i\sqrt{39}.

$$x=\frac{5+\sqrt{39}i}{4}$$

Now solve the equation x=\frac{5±\sqrt{39}i}{4} when ± is minus. Subtract i\sqrt{39} from 5.

$$x=\frac{-\sqrt{39}i+5}{4}$$

The equation is now solved.

$$x=\frac{5+\sqrt{39}i}{4} x=\frac{-\sqrt{39}i+5}{4}$$

Steps for Completing the Square

Subtract 5x from both sides.

$$2x^{2}-5x=-8$$

Divide both sides by 2.

$$\frac{2x^{2}-5x}{2}=-\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{5}{2}x=-\frac{8}{2}$$

Divide -8 by 2.

$$x^{2}-\frac{5}{2}x=-4$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-4+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-4+\frac{25}{16}$$

Add -4 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{39}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=-\frac{39}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{-\frac{39}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{39}i}{4} x-\frac{5}{4}=-\frac{\sqrt{39}i}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{5+\sqrt{39}i}{4} x=\frac{-\sqrt{39}i+5}{4}$$