<s>[INST] How to convert a quadratic solution to an unusual format I'm looking at old past papers and found this question:

"Solve the quadratic equation $3x^2 + 4x - 5$ giving your answer in the form $\frac{a}{b\pm\sqrt{19}}$, where $a$ and $b$ are integers."

I've never seen a quadratic solution in this form, with the surd root on the bottom. Does anybody have any hints on how to rearrange into this format?
Edit: For clarity, I've got $x = \frac{-2 \pm \sqrt{19}}{3}$, I just can't figure out how to convert that answer into the form they want.
 [/INST] Another approach would be to apply Viete's relations:  the sum of the roots of $ \ 3x^2 + 4x - 5 \ = \ 0 \ $ is
$$ \frac43 \ \ = \  \ - \left(\frac{a}{b + \sqrt{19}} \ + \ \frac{a}{b - \sqrt{19}} \right) \ \ =  \ \ -\frac{2ab}{b^2 - 19} $$
and the product of the roots is
$$ \frac{-5}{3} \ \ = \  \  \frac{a}{b + \sqrt{19}} \ · \ \frac{a}{b - \sqrt{19}}   \ \ =  \ \  \frac{a^2}{b^2 - 19} \ \ . $$
The ratio of these equations produces
$$ \frac{4/3}{-5/3} \ \ = \ \ -\frac45 \ \ = \ \ \frac{-2ab}{a^2} \ \ = \ \ -\frac{2b}{a} \ \ \Rightarrow \ \ a \ = \ \frac52 · b \ \ . $$
Inserting this into, say, the first equation  will lead to $  \ 19b^2 \ = \ 76 \ \ ; $ using the second equation yields a similar result. It will suffice to use the positive value for $ \ b \ \ , $ as using the negative value just gives an equivalent pair of roots in $ \ \frac{a}{b \ \pm \ \sqrt{19}} \ \ . $
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