Solve x=x+3/x-2

$$x= \frac{ x+3 }{ x-2 }$$

Solve for x

$$x = \frac{\sqrt{21} + 3}{2} \approx 3.791287847$$

$$x=\frac{3-\sqrt{21}}{2}\approx -0.791287847$$

Steps Using the Quadratic Formula

Subtract \frac{x+3}{x-2} from both sides.

$$x-\frac{x+3}{x-2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-2}{x-2}.

$$\frac{x\left(x-2\right)}{x-2}-\frac{x+3}{x-2}=0$$

Since \frac{x\left(x-2\right)}{x-2} and \frac{x+3}{x-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-2\right)-\left(x+3\right)}{x-2}=0$$

Do the multiplications in x\left(x-2\right)-\left(x+3\right).

$$\frac{x^{2}-2x-x-3}{x-2}=0$$

Combine like terms in x^{2}-2x-x-3.

$$\frac{x^{2}-3x-3}{x-2}=0$$

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x^{2}-3x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-3\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$x=\frac{-\left(-3\right)±\sqrt{9+12}}{2}$$

Add 9 to 12.

$$x=\frac{-\left(-3\right)±\sqrt{21}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{21}}{2}$$

Now solve the equation x=\frac{3±\sqrt{21}}{2} when ± is plus. Add 3 to \sqrt{21}.

$$x=\frac{\sqrt{21}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{21}}{2} when ± is minus. Subtract \sqrt{21} from 3.

$$x=\frac{3-\sqrt{21}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{21}+3}{2} x=\frac{3-\sqrt{21}}{2}$$

Steps for Completing the Square

Subtract \frac{x+3}{x-2} from both sides.

$$x-\frac{x+3}{x-2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x-2}{x-2}.

$$\frac{x\left(x-2\right)}{x-2}-\frac{x+3}{x-2}=0$$

Since \frac{x\left(x-2\right)}{x-2} and \frac{x+3}{x-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-2\right)-\left(x+3\right)}{x-2}=0$$

Do the multiplications in x\left(x-2\right)-\left(x+3\right).

$$\frac{x^{2}-2x-x-3}{x-2}=0$$

Combine like terms in x^{2}-2x-x-3.

$$\frac{x^{2}-3x-3}{x-2}=0$$

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x^{2}-3x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$x^{2}-3x=3$$

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 3 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{21}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{21}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{21}}{2} x-\frac{3}{2}=-\frac{\sqrt{21}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{21}+3}{2} x=\frac{3-\sqrt{21}}{2}$$