Solve {l}{18/text{febre}}{8x^2=5x-2}

$$\left. \begin{array} { l } { 18 / \text { febre } } \\ { 8 x ^ { 2 } = 5 x - 2 } \end{array} \right.$$

Solve for x

$$x=\frac{5+\sqrt{39}i}{16}\approx 0.3125+0.390312375i$$

$$x=\frac{-\sqrt{39}i+5}{16}\approx 0.3125-0.390312375i$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

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

Add 2 to both sides.

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

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

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

Square -5.

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

Multiply -4 times 8.

$$x=\frac{-\left(-5\right)±\sqrt{25-32\times 2}}{2\times 8}$$

Multiply -32 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-64}}{2\times 8}$$

Add 25 to -64.

$$x=\frac{-\left(-5\right)±\sqrt{-39}}{2\times 8}$$

Take the square root of -39.

$$x=\frac{-\left(-5\right)±\sqrt{39}i}{2\times 8}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{39}i}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{5±\sqrt{39}i}{16}$$

Now solve the equation x=\frac{5±\sqrt{39}i}{16} when ± is plus. Add 5 to i\sqrt{39}.

$$x=\frac{5+\sqrt{39}i}{16}$$

Now solve the equation x=\frac{5±\sqrt{39}i}{16} when ± is minus. Subtract i\sqrt{39} from 5.

$$x=\frac{-\sqrt{39}i+5}{16}$$

The equation is now solved.

$$x=\frac{5+\sqrt{39}i}{16} x=\frac{-\sqrt{39}i+5}{16}$$

Steps for Completing the Square

Subtract 5x from both sides.

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

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

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

$$x^{2}-\frac{5}{8}x=-\frac{1}{4}$$

Divide -\frac{5}{8}, the coefficient of the x term, by 2 to get -\frac{5}{16}. Then add the square of -\frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{8}x+\left(-\frac{5}{16}\right)^{2}=-\frac{1}{4}+\left(-\frac{5}{16}\right)^{2}$$

Square -\frac{5}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{1}{4}+\frac{25}{256}$$

Add -\frac{1}{4} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{39}{256}$$

Factor x^{2}-\frac{5}{8}x+\frac{25}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{16}\right)^{2}=-\frac{39}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{16}\right)^{2}}=\sqrt{-\frac{39}{256}}$$

Simplify.

$$x-\frac{5}{16}=\frac{\sqrt{39}i}{16} x-\frac{5}{16}=-\frac{\sqrt{39}i}{16}$$

Add \frac{5}{16} to both sides of the equation.

$$x=\frac{5+\sqrt{39}i}{16} x=\frac{-\sqrt{39}i+5}{16}$$