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(1-2x-3x²) / (3x-x²-5) >0 ;then what is the value of x?

$Given \frac{1 - 2 x - 3 x^{2}}{3 x - x^{2} - 5} > 0 \Rightarrow 1 - 2 x - 3 x^{2} > 0 \Rightarrow 3 x^{2} + 2 x - 1 < 0 By using qudratic formula we get x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \Rightarrow x = \frac{- 2 \pm \sqrt{(2)^{2} - 4 (3) (- 1)}}{2 \times 3} \Rightarrow x = \frac{- 2 \pm \sqrt{4 + 12}}{6} \Rightarrow x = \frac{- 2 \pm 4}{6} = - 1, . \frac{1}{3} \Rightarrow x < - 1, \frac{1}{3}$

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