Solve pi(3p-1)^2=2pi(r+2)

$$\pi ( 3 p - 1 ) ^ { 2 } = 2 \pi ( r + 2 )$$

Solve for r

$$r=\frac{3\left(p-1\right)\left(3p+1\right)}{2}$$

Steps for Solving Linear Equation

Cancel out \pi on both sides.

$$\left(3p-1\right)^{2}=2\left(r+2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3p-1\right)^{2}.

$$9p^{2}-6p+1=2\left(r+2\right)$$

Use the distributive property to multiply 2 by r+2.

$$9p^{2}-6p+1=2r+4$$

Swap sides so that all variable terms are on the left hand side.

$$2r+4=9p^{2}-6p+1$$

Subtract 4 from both sides.

$$2r=9p^{2}-6p+1-4$$

Subtract 4 from 1 to get -3.

$$2r=9p^{2}-6p-3$$

Divide both sides by 2.

$$\frac{2r}{2}=\frac{3\left(p-1\right)\left(3p+1\right)}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$r=\frac{3\left(p-1\right)\left(3p+1\right)}{2}$$

Solve for p

$$p=\frac{-\sqrt{2r+4}+1}{3}$$

$$p=\frac{\sqrt{2r+4}+1}{3}\text{, }r\geq -2$$