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(D) For equation $I$: $5x^2 + 3x - 14 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-3 \pm \sqrt{3^2 - 4(5)(-14)

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if $x = y$ or relationship between $x$ and $y$ cannot be established.

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(D) For equation $I$: $5x^2 + 3x - 14 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-3 \pm \sqrt{3^2 - 4(5)(-14)}}{2(5)} = \frac{-3 \pm \sqrt{9 + 280}}{10} = \frac{-3 \pm \sqrt{289}}{10} = \frac{-3 \pm 17}{10}$$x_1 = \frac{14}{10} = 1.4$ and $x_2 = \frac{-20}{10} = -2$.For equation $II$: $10y^2 - 3y - 27 = 0$Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$y = \frac{3 \pm \sqrt{(-3)^2 - 4(10)(-27)}}{2(10)} = \frac{3 \pm \sqrt{9 + 1080}}{20} = \frac{3 \pm \sqrt{1089}}{20} = \frac{3 \pm 33}{20}$$y_1 = \frac{36}{20} = 1.8$ and $y_2 = \frac{-30}{20} = -1.5$.Comparing the values:$x = \{1.4, -2\}$ and $y = \{1.8, -1.5\}$.Since $1.4  -1.5$,the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I.$ $5x^2 + 3x - 14 = 0$
$II.$ $10y^2 - 3y - 27 = 0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $5x^2 + 3x - 14 = 0$$II.$ $10y^2 - 3y - 27 = 0$