What is the inverse of 2/(s^2+3s-4) ?

The inverse of 2/(s^2+3s-4) is (-3s+sqrt(25s^2+8s))/(2s),(-3s-sqrt(25s^2+8s))/(2s)

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inverse of 2/(s^2+3s-4)

inverse

\frac{2}{s ^{2} + 3 s - 4}

\frac{- 3 s + 25 s ^{2} + 8 s}{2 s}, \frac{- 3 s - 25 s ^{2} + 8 s}{2 s}

Solution steps

y = \frac{2}{s ^{2} + 3 s - 4}

Replace

s

with

y

s = \frac{2}{y ^{2} + 3 y - 4}

Solve for

y, s = \frac{2}{y ^{2} + 3 y - 4}

\frac{- 3 s + 25 s ^{2} + 8 s}{2 s}, \frac{- 3 s - 25 s ^{2} + 8 s}{2 s}

z^{2} - 0.25 z - 0.375

f (x) = \frac{2}{7} x + 6

f (x) = - x - 8

f (x) = 1.5 - (\frac{1}{( 0.15 x + 1 )})

y = 2 9^{x}

What is the inverse of 2/(s^2+3s-4) ?
The inverse of 2/(s^2+3s-4) is (-3s+sqrt(25s^2+8s))/(2s),(-3s-sqrt(25s^2+8s))/(2s)