What is the inverse of ((1-s))/((s^2-4s+13)) ?

The inverse of ((1-s))/((s^2-4s+13)) is (4s-1+sqrt(-36s^2-4s+1))/(2s),(4s-1-sqrt(-36s^2-4s+1))/(2s)

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inverse of ((1-s))/((s^2-4s+13))

inverse

\frac{( 1 - s )}{( s ^{2} - 4 s + 13 )}

\frac{4 s - 1 + - 36 s ^{2} - 4 s + 1}{2 s}, \frac{4 s - 1 - - 36 s ^{2} - 4 s + 1}{2 s}

Solution steps

y = \frac{1 - s}{s ^{2} - 4 s + 13}

Replace

s

with

y

s = \frac{1 - y}{y ^{2} - 4 y + 13}

Solve for

y, s = \frac{1 - y}{y ^{2} - 4 y + 13}

\frac{4 s - 1 + - 36 s ^{2} - 4 s + 1}{2 s}, \frac{4 s - 1 - - 36 s ^{2} - 4 s + 1}{2 s}

y = \frac{1}{s ^{2} - s - 2}

g (x) = \frac{x}{3} + 6

7 x^{4} - 6 x^{2} - 1

f (x) = - 2^{3}

\frac{1}{x - 2} + 1

What is the inverse of ((1-s))/((s^2-4s+13)) ?
The inverse of ((1-s))/((s^2-4s+13)) is (4s-1+sqrt(-36s^2-4s+1))/(2s),(4s-1-sqrt(-36s^2-4s+1))/(2s)