What is the inverse of (2s+3)/(s(2s+1)) ?

The inverse of (2s+3)/(s(2s+1)) is (-s+2+sqrt(s^2+20s+4))/(4s),(-s+2-sqrt(s^2+20s+4))/(4s)

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inverse of (2s+3)/(s(2s+1))

inverse

\frac{2 s + 3}{s ( 2 s + 1 )}

\frac{- s + 2 + s ^{2} + 20 s + 4}{4 s}, \frac{- s + 2 - s ^{2} + 20 s + 4}{4 s}

Solution steps

y = \frac{2 s + 3}{s ( 2 s + 1 )}

Replace

s

with

y

s = \frac{2 y + 3}{y ( 2 y + 1 )}

Solve for

y, s = \frac{2 y + 3}{y ( 2 y + 1 )}

\frac{- s + 2 + s ^{2} + 20 s + 4}{4 s}, \frac{- s + 2 - s ^{2} + 20 s + 4}{4 s}

f (x) = \frac{1}{7} x^{2} + 3

f (x) = x^{2 - 10 + 15}

+ \frac{z ^{- 1}}{( 1 - 0.5 \cdot z ^{- 1} ) ^{2}}

y = \frac{1}{2 x - 4}

f (x) = \frac{x ^{2} + 3 x - 10}{x - 2}

What is the inverse of (2s+3)/(s(2s+1)) ?
The inverse of (2s+3)/(s(2s+1)) is (-s+2+sqrt(s^2+20s+4))/(4s),(-s+2-sqrt(s^2+20s+4))/(4s)