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

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

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inverse of 2pir^2+8pir

inverse

2 π r^{2} + 8 π r

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

Solution steps

y = 2 π r^{2} + 8 π r

Replace

r

with

y

r = 2 π y^{2} + 8 π y

Solve for

y, r = 2 π y^{2} + 8 π y

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

f (x) = 2 π x^{2} + 12 π x

f (x) = 2 - \frac{x ^{2}}{2}

f (x) = - 2 (x - 3)^{2} - 6

\frac{3 - 2 x}{2 x - 5}

f (x) = 2 - 3 x^{3}

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