Solve 0=8x^2+5-6x

$$0 = 8 { x }^{ 2 } +5-6x$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{31}i}{8}\approx 0.375+0.695970545i$$

$$x=\frac{-\sqrt{31}i+3}{8}\approx 0.375-0.695970545i$$

Steps Using the Quadratic Formula

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

$$8x^{2}+5-6x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$8x^{2}-6x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8\times 5}}{2\times 8}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8\times 5}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-6\right)±\sqrt{36-32\times 5}}{2\times 8}$$

Multiply -32 times 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-160}}{2\times 8}$$

Add 36 to -160.

$$x=\frac{-\left(-6\right)±\sqrt{-124}}{2\times 8}$$

Take the square root of -124.

$$x=\frac{-\left(-6\right)±2\sqrt{31}i}{2\times 8}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{31}i}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{6±2\sqrt{31}i}{16}$$

Now solve the equation x=\frac{6±2\sqrt{31}i}{16} when ± is plus. Add 6 to 2i\sqrt{31}.

$$x=\frac{6+2\sqrt{31}i}{16}$$

Divide 6+2i\sqrt{31} by 16.

$$x=\frac{3+\sqrt{31}i}{8}$$

Now solve the equation x=\frac{6±2\sqrt{31}i}{16} when ± is minus. Subtract 2i\sqrt{31} from 6.

$$x=\frac{-2\sqrt{31}i+6}{16}$$

Divide 6-2i\sqrt{31} by 16.

$$x=\frac{-\sqrt{31}i+3}{8}$$

The equation is now solved.

$$x=\frac{3+\sqrt{31}i}{8} x=\frac{-\sqrt{31}i+3}{8}$$

Steps for Completing the Square

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

$$8x^{2}+5-6x=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}-6x=-5$$

Divide both sides by 8.

$$\frac{8x^{2}-6x}{8}=-\frac{5}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{6}{8}\right)x=-\frac{5}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{4}x=-\frac{5}{8}$$

Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=-\frac{5}{8}+\left(-\frac{3}{8}\right)^{2}$$

Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{5}{8}+\frac{9}{64}$$

Add -\frac{5}{8} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{31}{64}$$

Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{8}\right)^{2}=-\frac{31}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{-\frac{31}{64}}$$

Simplify.

$$x-\frac{3}{8}=\frac{\sqrt{31}i}{8} x-\frac{3}{8}=-\frac{\sqrt{31}i}{8}$$

Add \frac{3}{8} to both sides of the equation.

$$x=\frac{3+\sqrt{31}i}{8} x=\frac{-\sqrt{31}i+3}{8}$$