Solve x=5-(2divx)

$$x=5-(2 \div x)$$

Solve for x

$$x = \frac{\sqrt{17} + 5}{2} \approx 4.561552813$$

$$x=\frac{5-\sqrt{17}}{2}\approx 0.438447187$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{x}{x}.

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

Since \frac{5x}{x} and \frac{2}{x} have the same denominator, subtract them by subtracting their numerators.

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

Subtract \frac{5x-2}{x} from both sides.

$$x-\frac{5x-2}{x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x}{x}.

$$\frac{xx}{x}-\frac{5x-2}{x}=0$$

Since \frac{xx}{x} and \frac{5x-2}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{xx-\left(5x-2\right)}{x}=0$$

Do the multiplications in xx-\left(5x-2\right).

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

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

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

Multiply -4 times 2.

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

Add 25 to -8.

$$x=\frac{-\left(-5\right)±\sqrt{17}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{17}}{2}$$

Now solve the equation x=\frac{5±\sqrt{17}}{2} when ± is plus. Add 5 to \sqrt{17}.

$$x=\frac{\sqrt{17}+5}{2}$$

Now solve the equation x=\frac{5±\sqrt{17}}{2} when ± is minus. Subtract \sqrt{17} from 5.

$$x=\frac{5-\sqrt{17}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+5}{2} x=\frac{5-\sqrt{17}}{2}$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{x}{x}.

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

Since \frac{5x}{x} and \frac{2}{x} have the same denominator, subtract them by subtracting their numerators.

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

Subtract \frac{5x-2}{x} from both sides.

$$x-\frac{5x-2}{x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x}{x}.

$$\frac{xx}{x}-\frac{5x-2}{x}=0$$

Since \frac{xx}{x} and \frac{5x-2}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{xx-\left(5x-2\right)}{x}=0$$

Do the multiplications in xx-\left(5x-2\right).

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

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

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -2 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{17}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{17}}{2} x-\frac{5}{2}=-\frac{\sqrt{17}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{17}+5}{2} x=\frac{5-\sqrt{17}}{2}$$