What is the inverse of 2pir^2+8pirh ?

The inverse of 2pir^2+8pirh is (-8pih+sqrt(64pi^2h^2+8pir))/(4pi),(-8pih-sqrt(64pi^2h^2+8pir))/(4pi)

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inverse of 2pir^2+8pirh

inverse

2 π r^{2} + 8 π r h

\frac{- 8 πh + 64 π ^{2} h ^{2} + 8 π r}{4 π}, \frac{- 8 πh - 64 π ^{2} h ^{2} + 8 π r}{4 π}

Solution steps

y = 2 π r^{2} + 8 π r h

Replace

r

with

y

r = 2 π y^{2} + 8 π y h

Solve for

y, r = 2 π y^{2} + 8 π y h

\frac{- 8 πh + 64 π ^{2} h ^{2} + 8 π r}{4 π}, \frac{- 8 πh - 64 π ^{2} h ^{2} + 8 π r}{4 π}

What is the inverse of 2pir^2+8pirh ?
The inverse of 2pir^2+8pirh is (-8pih+sqrt(64pi^2h^2+8pir))/(4pi),(-8pih-sqrt(64pi^2h^2+8pir))/(4pi)