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(D) For equation $I$: $12 x^{2} + 11 x - 56 = 0$.Splitting the middle term: $12 x^{2} + 32 x - 21 x - 56 = 0$.$4 x(3 x + 8) - 7(3 x + 8) = 0$.$(4 x -

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if $x \le y$

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(D) For equation $I$: $12 x^{2} + 11 x - 56 = 0$.Splitting the middle term: $12 x^{2} + 32 x - 21 x - 56 = 0$.$4 x(3 x + 8) - 7(3 x + 8) = 0$.$(4 x - 7)(3 x + 8) = 0$.So,$x = \frac{7}{4} = 1.75$ or $x = -\frac{8}{3} \approx -2.67$.For equation $II$: $4 y^{2} - 15 y + 14 = 0$.Splitting the middle term: $4 y^{2} - 8 y - 7 y + 14 = 0$.$4 y(y - 2) - 7(y - 2) = 0$.$(4 y - 7)(y - 2) = 0$.So,$y = \frac{7}{4} = 1.75$ or $y = 2$.Comparing the values:$x = \{1.75, -2.67\}$ and $y = \{1.75, 2\}$.Since $1.75 \le 1.75$,$1.75 < 2$,$-2.67 < 1.75$,and $-2.67 < 2$,we conclude that $x \le y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $12 x^{2} + 11 x - 56 = 0$
$II.$ $4 y^{2} - 15 y + 14 = 0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $12 x^{2} + 11 x - 56 = 0$$II.$ $4 y^{2} - 15 y + 14 = 0$