Solve 90=R+30R+150/R+35

$$90 = R + \frac { 30 R + 150 } { R + 35 }$$

Solve for R

$$R = \frac{5 \sqrt{505} + 25}{2} \approx 68.680512636$$

$$R=\frac{25-5\sqrt{505}}{2}\approx -43.680512636$$

Steps Using the Quadratic Formula

Variable R cannot be equal to -35 since division by zero is not defined. Multiply both sides of the equation by R+35.

$$90\left(R+35\right)=\left(R+35\right)R+30R+150$$

Use the distributive property to multiply 90 by R+35.

$$90R+3150=\left(R+35\right)R+30R+150$$

Use the distributive property to multiply R+35 by R.

$$90R+3150=R^{2}+35R+30R+150$$

Combine 35R and 30R to get 65R.

$$90R+3150=R^{2}+65R+150$$

Subtract R^{2} from both sides.

$$90R+3150-R^{2}=65R+150$$

Subtract 65R from both sides.

$$90R+3150-R^{2}-65R=150$$

Combine 90R and -65R to get 25R.

$$25R+3150-R^{2}=150$$

Subtract 150 from both sides.

$$25R+3150-R^{2}-150=0$$

Subtract 150 from 3150 to get 3000.

$$25R+3000-R^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-R^{2}+25R+3000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 25 for b, and 3000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$R=\frac{-25±\sqrt{25^{2}-4\left(-1\right)\times 3000}}{2\left(-1\right)}$$

Square 25.

$$R=\frac{-25±\sqrt{625-4\left(-1\right)\times 3000}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$R=\frac{-25±\sqrt{625+4\times 3000}}{2\left(-1\right)}$$

Multiply 4 times 3000.

$$R=\frac{-25±\sqrt{625+12000}}{2\left(-1\right)}$$

Add 625 to 12000.

$$R=\frac{-25±\sqrt{12625}}{2\left(-1\right)}$$

Take the square root of 12625.

$$R=\frac{-25±5\sqrt{505}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$R=\frac{-25±5\sqrt{505}}{-2}$$

Now solve the equation R=\frac{-25±5\sqrt{505}}{-2} when ± is plus. Add -25 to 5\sqrt{505}.

$$R=\frac{5\sqrt{505}-25}{-2}$$

Divide -25+5\sqrt{505} by -2.

$$R=\frac{25-5\sqrt{505}}{2}$$

Now solve the equation R=\frac{-25±5\sqrt{505}}{-2} when ± is minus. Subtract 5\sqrt{505} from -25.

$$R=\frac{-5\sqrt{505}-25}{-2}$$

Divide -25-5\sqrt{505} by -2.

$$R=\frac{5\sqrt{505}+25}{2}$$

The equation is now solved.

$$R=\frac{25-5\sqrt{505}}{2} R=\frac{5\sqrt{505}+25}{2}$$

Steps for Completing the Square

Variable R cannot be equal to -35 since division by zero is not defined. Multiply both sides of the equation by R+35.

$$90\left(R+35\right)=\left(R+35\right)R+30R+150$$

Use the distributive property to multiply 90 by R+35.

$$90R+3150=\left(R+35\right)R+30R+150$$

Use the distributive property to multiply R+35 by R.

$$90R+3150=R^{2}+35R+30R+150$$

Combine 35R and 30R to get 65R.

$$90R+3150=R^{2}+65R+150$$

Subtract R^{2} from both sides.

$$90R+3150-R^{2}=65R+150$$

Subtract 65R from both sides.

$$90R+3150-R^{2}-65R=150$$

Combine 90R and -65R to get 25R.

$$25R+3150-R^{2}=150$$

Subtract 3150 from both sides.

$$25R-R^{2}=150-3150$$

Subtract 3150 from 150 to get -3000.

$$25R-R^{2}=-3000$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-R^{2}+25R=-3000$$

Divide both sides by -1.

$$\frac{-R^{2}+25R}{-1}=-\frac{3000}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$R^{2}+\frac{25}{-1}R=-\frac{3000}{-1}$$

Divide 25 by -1.

$$R^{2}-25R=-\frac{3000}{-1}$$

Divide -3000 by -1.

$$R^{2}-25R=3000$$

Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$R^{2}-25R+\left(-\frac{25}{2}\right)^{2}=3000+\left(-\frac{25}{2}\right)^{2}$$

Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$R^{2}-25R+\frac{625}{4}=3000+\frac{625}{4}$$

Add 3000 to \frac{625}{4}.

$$R^{2}-25R+\frac{625}{4}=\frac{12625}{4}$$

Factor R^{2}-25R+\frac{625}{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}.

$$\left(R-\frac{25}{2}\right)^{2}=\frac{12625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(R-\frac{25}{2}\right)^{2}}=\sqrt{\frac{12625}{4}}$$

Simplify.

$$R-\frac{25}{2}=\frac{5\sqrt{505}}{2} R-\frac{25}{2}=-\frac{5\sqrt{505}}{2}$$

Add \frac{25}{2} to both sides of the equation.

$$R=\frac{5\sqrt{505}+25}{2} R=\frac{25-5\sqrt{505}}{2}$$