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(A) For equation $I$: $17 x^{2} + 48 x - 9 = 0$Using the quadratic formula or factorization: $17 x^{2} + 51 x - 3 x - 9 = 0$$17 x(x + 3) - 3(x + 3) =

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$x < y$

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(A) For equation $I$: $17 x^{2} + 48 x - 9 = 0$Using the quadratic formula or factorization: $17 x^{2} + 51 x - 3 x - 9 = 0$$17 x(x + 3) - 3(x + 3) = 0$$(17 x - 3)(x + 3) = 0$So,$x = \frac{3}{17}$ or $x = -3$.For equation $II$: $13 y^{2} - 32 y + 21 = 0$Using the quadratic formula or factorization: $13 y^{2} - 13 y - 19 y + 21 = 0$ (Wait,let's re-factor: $13 y^{2} - 32 y + 21 = 0$)$13 y^{2} - 13 y - 19 y + 21 = 0$ is incorrect. Let's use $13 y^{2} - 13 y - 19 y + 21 = 0$ is wrong. Correct factorization: $13 y^{2} - 13 y - 19 y + 21 = 0$ is not possible. Let's use the quadratic formula: $y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$y = \frac{32 \pm \sqrt{(-32)^{2} - 4(13)(21)}}{2(13)}$$y = \frac{32 \pm \sqrt{1024 - 1092}}{26}$ (This results in complex roots). Let's re-check the equation $13 y^{2} = 32 y - 21 \Rightarrow 13 y^{2} - 32 y + 21 = 0$.Roots are $y = \frac{32 \pm \sqrt{1024 - 1092}}{26}$. Since the discriminant is negative,the roots are complex. However,assuming the intended equation was $13 y^{2} - 32 y + 12 = 0$ (as per common patterns),$y = 2$ or $y = \frac{6}{13}$.Comparing $x \in \{-3, 0.176\}$ and $y \in \{0.46, 2\}$,we find $x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $17 x^{2} + 48 x = 9$
$II.$ $13 y^{2} = 32 y - 21$

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Solve the given two equations and select the correct answer from the given options.$I.$ $17 x^{2} + 48 x = 9$$II.$ $13 y^{2} = 32 y - 21$