Solve R=2+R/(R+1)

$$R = 2 + R / ( R + 1 )$$

Solve for R

$$R=\sqrt{3}+1\approx 2.732050808$$

$$R=1-\sqrt{3}\approx -0.732050808$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{R+1}{R+1}.

$$R=\frac{2\left(R+1\right)}{R+1}+\frac{R}{R+1}$$

Since \frac{2\left(R+1\right)}{R+1} and \frac{R}{R+1} have the same denominator, add them by adding their numerators.

$$R=\frac{2\left(R+1\right)+R}{R+1}$$

Do the multiplications in 2\left(R+1\right)+R.

$$R=\frac{2R+2+R}{R+1}$$

Combine like terms in 2R+2+R.

$$R=\frac{3R+2}{R+1}$$

Subtract \frac{3R+2}{R+1} from both sides.

$$R-\frac{3R+2}{R+1}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply R times \frac{R+1}{R+1}.

$$\frac{R\left(R+1\right)}{R+1}-\frac{3R+2}{R+1}=0$$

Since \frac{R\left(R+1\right)}{R+1} and \frac{3R+2}{R+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{R\left(R+1\right)-\left(3R+2\right)}{R+1}=0$$

Do the multiplications in R\left(R+1\right)-\left(3R+2\right).

$$\frac{R^{2}+R-3R-2}{R+1}=0$$

Combine like terms in R^{2}+R-3R-2.

$$\frac{R^{2}-2R-2}{R+1}=0$$

Variable R cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by R+1.

$$R^{2}-2R-2=0$$

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

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

Square -2.

$$R=\frac{-\left(-2\right)±\sqrt{4-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$R=\frac{-\left(-2\right)±\sqrt{4+8}}{2}$$

Add 4 to 8.

$$R=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

Take the square root of 12.

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

The opposite of -2 is 2.

$$R=\frac{2±2\sqrt{3}}{2}$$

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

$$R=\frac{2\sqrt{3}+2}{2}$$

Divide 2+2\sqrt{3} by 2.

$$R=\sqrt{3}+1$$

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

$$R=\frac{2-2\sqrt{3}}{2}$$

Divide 2-2\sqrt{3} by 2.

$$R=1-\sqrt{3}$$

The equation is now solved.

$$R=\sqrt{3}+1 R=1-\sqrt{3}$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{R+1}{R+1}.

$$R=\frac{2\left(R+1\right)}{R+1}+\frac{R}{R+1}$$

Since \frac{2\left(R+1\right)}{R+1} and \frac{R}{R+1} have the same denominator, add them by adding their numerators.

$$R=\frac{2\left(R+1\right)+R}{R+1}$$

Do the multiplications in 2\left(R+1\right)+R.

$$R=\frac{2R+2+R}{R+1}$$

Combine like terms in 2R+2+R.

$$R=\frac{3R+2}{R+1}$$

Subtract \frac{3R+2}{R+1} from both sides.

$$R-\frac{3R+2}{R+1}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply R times \frac{R+1}{R+1}.

$$\frac{R\left(R+1\right)}{R+1}-\frac{3R+2}{R+1}=0$$

Since \frac{R\left(R+1\right)}{R+1} and \frac{3R+2}{R+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{R\left(R+1\right)-\left(3R+2\right)}{R+1}=0$$

Do the multiplications in R\left(R+1\right)-\left(3R+2\right).

$$\frac{R^{2}+R-3R-2}{R+1}=0$$

Combine like terms in R^{2}+R-3R-2.

$$\frac{R^{2}-2R-2}{R+1}=0$$

Variable R cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by R+1.

$$R^{2}-2R-2=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$R^{2}-2R=2$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$R^{2}-2R+1=2+1$$

Add 2 to 1.

$$R^{2}-2R+1=3$$

Factor R^{2}-2R+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(R-1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(R-1\right)^{2}}=\sqrt{3}$$

Simplify.

$$R-1=\sqrt{3} R-1=-\sqrt{3}$$

Add 1 to both sides of the equation.

$$R=\sqrt{3}+1 R=1-\sqrt{3}$$