The reciprocals of the roots $\rm\,r,s,1/2\:$ are roots of the reversed polynomial $\rm\:2\, x^3\! -\! 3\, x^2\! -\! 23\, x\! +\! 12.\: $ Thus by Vieta's Formulas $\rm\:r\!+\!s\!+\!1/2 = 3/2,\ rs/2 = -6,\:$ so $\rm\:r\!+\!s = 1,\ rs = -12,\:$ so $\rm\:r,s = \ldots$
Alternatively, by the Factor Theorem, $\rm\:f(2) = 0\:$ $\Rightarrow$ $\rm\:f(x)\:$ has $\rm\:x\!-\!2\:$ as a factor. Comparing coef's
$\rm (x-2)(a\, x^2 + b\, x + c)\ =\ 12\, x^3 - 23\,x^2 -3\,x + 2$
$\rm\qquad x^3\:$ coef $\rm\:\Rightarrow\: a = 12$
$\rm\qquad x^0\:$ coef $\rm\:\Rightarrow\: -2\,c = 2\:\Rightarrow\: c = -1$
$\rm\qquad x^1\:$ coef $\rm\:\Rightarrow\: -3 = c-2b = -1-2b\:\Rightarrow\: b = 1 $
So the quadratic factor is $\rm\: 12\,x^2 + x - 1,\:$ which can be solved by either the Quadratic Formula or Rational Root Test, or $\rm\:(c\,x\!-\!1)\,(d\,x\!+\!1) = 12\,x^2\!+\!x\!-\!1\:$ $\Rightarrow$ $\rm\:cd=12,\ c\!-\!d = 1,\:$ so $\rm\:c,d = \ldots$