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Решите задачу:

$\frac{3p^{2}+p-2}{4-9p^{2}} \\ \\ 3p^{2}+p-2=0 \\ D=1-4*3*(-2)=25=5^{2} \\ p_{1}= \frac{-1+5}{6}= \frac{2}{3} \\ p_{2}= \frac{-1-5}{6}=-1 \\ 3p^{2}+p-2=3(p- \frac{2}{3})(p+1) \\ \\ \frac{3p^{2}+p-2}{4-9p^{2}}= \frac{3(p- \frac{2}{3} )(p+1)}{(2-3p)(2+3p)}= \frac{(3p-2)(p+1)}{(2-3p)(2+3p)}= -\frac{(3p-2)(p+1)}{(3p-2)(3p+2)} = -\frac{p+1}{3p+2}$