Solve x=x+2-16/x-4

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

Solve for x (complex solution)

$$x=\frac{5+\sqrt{31}i}{2}\approx 2.5+2.783882181i$$

$$x=\frac{-\sqrt{31}i+5}{2}\approx 2.5-2.783882181i$$

Steps Using the Quadratic Formula

Subtract 16 from 2 to get -14.

$$x=\frac{x-14}{x-4}$$

Subtract \frac{x-14}{x-4} from both sides.

$$x-\frac{x-14}{x-4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-4}{x-4}.

$$\frac{x\left(x-4\right)}{x-4}-\frac{x-14}{x-4}=0$$

Since \frac{x\left(x-4\right)}{x-4} and \frac{x-14}{x-4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-4\right)-\left(x-14\right)}{x-4}=0$$

Do the multiplications in x\left(x-4\right)-\left(x-14\right).

$$\frac{x^{2}-4x-x+14}{x-4}=0$$

Combine like terms in x^{2}-4x-x+14.

$$\frac{x^{2}-5x+14}{x-4}=0$$

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

$$x^{2}-5x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 14}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 14}}{2}$$

Multiply -4 times 14.

$$x=\frac{-\left(-5\right)±\sqrt{25-56}}{2}$$

Add 25 to -56.

$$x=\frac{-\left(-5\right)±\sqrt{-31}}{2}$$

Take the square root of -31.

$$x=\frac{-\left(-5\right)±\sqrt{31}i}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{31}i}{2}$$

Now solve the equation x=\frac{5±\sqrt{31}i}{2} when ± is plus. Add 5 to i\sqrt{31}.

$$x=\frac{5+\sqrt{31}i}{2}$$

Now solve the equation x=\frac{5±\sqrt{31}i}{2} when ± is minus. Subtract i\sqrt{31} from 5.

$$x=\frac{-\sqrt{31}i+5}{2}$$

The equation is now solved.

$$x=\frac{5+\sqrt{31}i}{2} x=\frac{-\sqrt{31}i+5}{2}$$

Steps for Completing the Square

Subtract 16 from 2 to get -14.

$$x=\frac{x-14}{x-4}$$

Subtract \frac{x-14}{x-4} from both sides.

$$x-\frac{x-14}{x-4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-4}{x-4}.

$$\frac{x\left(x-4\right)}{x-4}-\frac{x-14}{x-4}=0$$

Since \frac{x\left(x-4\right)}{x-4} and \frac{x-14}{x-4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-4\right)-\left(x-14\right)}{x-4}=0$$

Do the multiplications in x\left(x-4\right)-\left(x-14\right).

$$\frac{x^{2}-4x-x+14}{x-4}=0$$

Combine like terms in x^{2}-4x-x+14.

$$\frac{x^{2}-5x+14}{x-4}=0$$

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by x-4.

$$x^{2}-5x+14=0$$

Subtract 14 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-5x=-14$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-14+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-14+\frac{25}{4}$$

Add -14 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=-\frac{31}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{31}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{31}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{31}i}{2} x-\frac{5}{2}=-\frac{\sqrt{31}i}{2}$$

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

$$x=\frac{5+\sqrt{31}i}{2} x=\frac{-\sqrt{31}i+5}{2}$$