What is the inverse of (3-s)/(s^2-4s+7) ?

The inverse of (3-s)/(s^2-4s+7) is (4s-1+sqrt(-12s^2+4s+1))/(2s),(4s-1-sqrt(-12s^2+4s+1))/(2s)

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inverse of (3-s)/(s^2-4s+7)

inverse

\frac{3 - s}{s ^{2} - 4 s + 7}

\frac{4 s - 1 + - 12 s ^{2} + 4 s + 1}{2 s}, \frac{4 s - 1 - - 12 s ^{2} + 4 s + 1}{2 s}

Solution steps

y = \frac{3 - s}{s ^{2} - 4 s + 7}

Replace

s

with

y

s = \frac{3 - y}{y ^{2} - 4 y + 7}

Solve for

y, s = \frac{3 - y}{y ^{2} - 4 y + 7}

\frac{4 s - 1 + - 12 s ^{2} + 4 s + 1}{2 s}, \frac{4 s - 1 - - 12 s ^{2} + 4 s + 1}{2 s}

\frac{4 x ^{2} - 256}{x + 8}

f (x) = 4 x - 3 - 5

4 x^{2} + 34 x

7 - 4 x^{3}

f (L) = 124.5 + 45 L - 0.03 L^{2}

What is the inverse of (3-s)/(s^2-4s+7) ?
The inverse of (3-s)/(s^2-4s+7) is (4s-1+sqrt(-12s^2+4s+1))/(2s),(4s-1-sqrt(-12s^2+4s+1))/(2s)