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(B) For equation $I: 2x^2 + 3x + 1 = 0$Factoring the quadratic equation: $2x^2 + 2x + x + 1 = 0$$2x(x + 1) + 1(x + 1) = 0$$(2x + 1)(x + 1) = 0$Thus,$x

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if $x < y$

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(B) For equation $I: 2x^2 + 3x + 1 = 0$Factoring the quadratic equation: $2x^2 + 2x + x + 1 = 0$$2x(x + 1) + 1(x + 1) = 0$$(2x + 1)(x + 1) = 0$Thus,$x_1 = -1$ and $x_2 = -0.5$.For equation $II: 12y^2 + 7y + 1 = 0$Factoring the quadratic equation: $12y^2 + 4y + 3y + 1 = 0$$4y(3y + 1) + 1(3y + 1) = 0$$(4y + 1)(3y + 1) = 0$Thus,$y_1 = -1/4 = -0.25$ and $y_2 = -1/3 \approx -0.33$.Comparing the values:$x$ values are $\{-1, -0.5\}$$y$ values are $\{-0.25, -0.33\}$Since both values of $x$ are less than both values of $y$ (e.g.,$-0.5 < -0.33$),we conclude that $x < y$.

Solve the given two equations and select the correct answer from the given options.
$I. 2x^2 + 3x + 1 = 0$
$II. 12y^2 + 7y + 1 = 0$

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Solve the given two equations and select the correct answer from the given options.$I. 2x^2 + 3x + 1 = 0$$II. 12y^2 + 7y + 1 = 0$