# I am having trouble finding the inverse Laplace transform of: \frac{1}{s^2-9s+20} I

I am having trouble finding the inverse Laplace transform of:
$\frac{1}{{s}^{2}-9s+20}$
I tried writing it in a different way:
${L}^{-1}\left\{\frac{1}{{\left(s-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\right\}=2{L}^{-1}\left\{\frac{\frac{1}{2}}{{\left(s-\frac{9}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}}\right\}$
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Yaretzi Odom
I'll give you a hint, because I think it might be useful for you to get practice. If you need more tell in comment: try rewriting this using partial fractions and then use the linearity of the inverse laplace.
###### Not exactly what you’re looking for?
amisayq6t
I have done it:
$X\left(s\right)=\frac{1}{{s}^{2}-9s+20}=\frac{1}{\left(s-4\right)\left(s-5\right)}=\frac{-1}{s-4}+\frac{1}{s-5}$
$=>x\left(t\right)=-{e}^{4t}+{e}^{5t}$