What is the inverse of s^2+4.5s+9 ?

The inverse of s^2+4.5s+9 is (-9+sqrt(16s-63))/4 ,-(sqrt(16s-63)+9)/4

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inverse of s^2+4.5s+9

inverse

s^{2} + 4.5 s + 9

\frac{- 9 + 16 s - 63}{4}, - \frac{16 s - 63 + 9}{4}

Solution steps

y = s^{2} + 4.5 s + 9

Replace

s

with

y

s = y^{2} + 4.5 y + 9

Solve for

y, s = y^{2} + 4.5 y + 9

\frac{- 9 + 16 s - 63}{4}, - \frac{16 s - 63 + 9}{4}

f (x) = cos (x - 1) + 2

- 2 x - 2

f (x) = e^{0.5 x}

f (x) = x^{(2)} - 8 x + 15

f (x) = [1.6] [2.8]

What is the inverse of s^2+4.5s+9 ?
The inverse of s^2+4.5s+9 is (-9+sqrt(16s-63))/4 ,-(sqrt(16s-63)+9)/4