Solve for r A(r)=2pir^2+8pir

$A (r) = 2 π r^{2} + 8 π r$

Since $r$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 π r^{2} + 8 π r = A r$

Subtract $A r$ from both sides of the equation.

$2 π r^{2} + 8 π r - A r = 0$

Factor $r$ out of $2 π r^{2} + 8 π r - A r$ .

Factor $r$ out of $2 π r^{2}$ .

$r (2 π r) + 8 π r - A r = 0$

Factor $r$ out of $8 π r$ .

$r (2 π r) + r (8 π) - A r = 0$

Factor $r$ out of $- A r$ .

$r (2 π r) + r (8 π) + r (- A) = 0$

Factor $r$ out of $r (2 π r) + r (8 π)$ .

$r (2 π r + 8 π) + r (- A) = 0$

Factor $r$ out of $r (2 π r + 8 π) + r (- A)$ .

$r (2 π r + 8 π - A) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$r = 0$

$2 π r + 8 π - A = 0$

Set $r$ equal to $0$ .

$r = 0$

Set $2 π r + 8 π - A$ equal to $0$ and solve for $r$ .

Set $2 π r + 8 π - A$ equal to $0$ .

$2 π r + 8 π - A = 0$

Solve $2 π r + 8 π - A = 0$ for $r$ .

Move all terms not containing $r$ to the right side of the equation.

Subtract $8 π$ from both sides of the equation.

$2 π r - A = - 8 π$

Add $A$ to both sides of the equation.

$2 π r = - 8 π + A$

Divide each term in $2 π r = - 8 π + A$ by $2 π$ and simplify.

Divide each term in $2 π r = - 8 π + A$ by $2 π$ .

$\frac{2 π r}{2 π} = \frac{- 8 π}{2 π} + \frac{A}{2 π}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 π r}{2 π} = \frac{- 8 π}{2 π} + \frac{A}{2 π}$

Rewrite the expression.

$\frac{π r}{π} = \frac{- 8 π}{2 π} + \frac{A}{2 π}$

Cancel the common factor of $π$ .

Cancel the common factor.

$\frac{π r}{π} = \frac{- 8 π}{2 π} + \frac{A}{2 π}$

Divide $r$ by $1$ .

$r = \frac{- 8 π}{2 π} + \frac{A}{2 π}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 8$ and $2$ .

Factor $2$ out of $- 8 π$ .

$r = \frac{2 (- 4 π)}{2 π} + \frac{A}{2 π}$

Cancel the common factors.

Factor $2$ out of $2 π$ .

$r = \frac{2 (- 4 π)}{2 (π)} + \frac{A}{2 π}$

Cancel the common factor.

$r = \frac{2 (- 4 π)}{2 π} + \frac{A}{2 π}$

Rewrite the expression.

$r = \frac{- 4 π}{π} + \frac{A}{2 π}$

Cancel the common factor of $π$ .

Cancel the common factor.

$r = \frac{- 4 π}{π} + \frac{A}{2 π}$

Divide $- 4$ by $1$ .

$r = - 4 + \frac{A}{2 π}$

The final solution is all the values that make $r (2 π r + 8 π - A) = 0$ true.

$r = 0$

$r = - 4 + \frac{A}{2 π}$

Solve for r A(r)=2pir^2+8pir

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