Calculate the fourier transform of $(ax^2+bx+c)^{-1}$

You may write $$ ax^2+bx+c = aleft(x-ifrac{sqrt{4ac-b^2}}{2a}right)left(x+ifrac{sqrt{4ac-b^2}}{2a}right). $$ Let $r=sqrt{4ac-b^2}/2a$, which is real and positive by assumption. Then $$ax^2+bx+c=a(x-ir)(x+ir)$$ and $$ frac{1}{ax^2+bx+c} = frac{1}{2iar}left(frac{1}{x-ir}-frac{1}{x+ir}right) $$ Then $$ hat{f}(xi)=frac{pi}{ar}frac{1}{2pi i}int_{-infty}^{infty}e^{2pi i zxi}left(frac{1}{z-ir}-frac{1}{z+ir}right)dz $$ If $xi > 0$, then $e^{2pi ixxi}$ decays as $Im zrightarrow infty$, and the integral may be traded for the limit of a positively-oriented, semicircular contour integral that closes in the upper half-plane, which may be evaluated using the residue at $ir$ to be $$ frac{pi}{ar}e^{-2pi rxi}. $$ If $xi < 0$, then $e^{2pi izxi}$ decays as $Im z rightarrow -infty$, and the integral may be traded for the limit of a negatively-oriented, semicircular contour integral that closes in the lower half-plane, which may be evaluated using the residue at $-ir$ to be $$ frac{pi}{ar}e^{2pi rxi} $$ Putting the two pieces together for positive and negative real $xi$ gives $$ hat{f}(xi) = frac{pi}{ar}e^{-2pi r|xi|},;;; r=sqrt{4ac-b^2}/2a $$

Completing the square, $$ ax^2 + bx + c = abigg( Big( x + frac{b}{2a} Big)^2 + frac{4ac-b^2}{4a^2} bigg) $$ For simplicity, suppose that the roots of the quadratic are not real, so we don't have to worry about singularities. Hence the second term in the bracket is positive, so it will suffice to find the Fourier transform of $ f(y) = 1/(y^2 + m^2) $ and apply various shift and scaling properties.

But we know that this is $$ mathcal{F}left( frac{1}{y^2+m^2} right)(k) = mathcal{F}^{-1}left( frac{1}{y^2+m^2} right)(k) = frac{pi}{m} e^{-2pi mlvert k rvert} $$ (the function is even, and the Fourier transform with the normalisation you specify is its own inverse for even functions).

Now, we want the transform of $ f(x+b/2)/a$ with $m= sqrt{4ac-b^2}/(2a)$. If $g(x) = f(x-alpha)$, $mathcal{F}(g)(k) = e^{-2pi i alpha}mathcal{F}(f)(k)$, so $$ mathcal{F}left( frac{1/a}{(x+b/2)^2+m^2} right)(k) = frac{2pi}{sqrt{4ac-b^2}} expleft( frac{pi}{a} big( i - sqrt{4ac-b^2}lvert k rvert big) right) $$