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(C) For equation $I$: $15 x^{2}-41 x+14=0$We factorize the quadratic equation: $15 x^{2}-35 x-6 x+14=0$$5 x(3 x-7)-2(3 x-7)=0$$(5 x-2)(3 x-7)=0$So,$x

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if $x < y$

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(C) For equation $I$: $15 x^{2}-41 x+14=0$We factorize the quadratic equation: $15 x^{2}-35 x-6 x+14=0$$5 x(3 x-7)-2(3 x-7)=0$$(5 x-2)(3 x-7)=0$So,$x = \frac{2}{5} = 0.4$ or $x = \frac{7}{3} \approx 2.33$.For equation $II$: $2 y^{2}-13 y+20=0$We factorize the quadratic equation: $2 y^{2}-8 y-5 y+20=0$$2 y(y-4)-5(y-4)=0$$(2 y-5)(y-4)=0$So,$y = \frac{5}{2} = 2.5$ or $y = 4$.Comparing the values:$x_1 = 0.4, x_2 = 2.33$$y_1 = 2.5, y_2 = 4$Since both values of $x$ are less than both values of $y$ $(0.4 < 2.5, 0.4 < 4, 2.33 < 2.5, 2.33 < 4)$,we conclude that $x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $15 x^{2}-41 x+14=0$
$II.$ $2 y^{2}-13 y+20=0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $15 x^{2}-41 x+14=0$$II.$ $2 y^{2}-13 y+20=0$