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(D) For equation $I$: $x^{2}+5x+6=0$Factorizing the quadratic equation: $x^{2}+3x+2x+6=0$$x(x+3)+2(x+3)=0$$(x+3)(x+2)=0$Therefore,the roots are $x_{1}

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if $x \le y$

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(D) For equation $I$: $x^{2}+5x+6=0$Factorizing the quadratic equation: $x^{2}+3x+2x+6=0$$x(x+3)+2(x+3)=0$$(x+3)(x+2)=0$Therefore,the roots are $x_{1}=-3$ and $x_{2}=-2$.For equation $II$: $y^{2}+3y+2=0$Factorizing the quadratic equation: $y^{2}+2y+y+2=0$$y(y+2)+1(y+2)=0$$(y+2)(y+1)=0$Therefore,the roots are $y_{1}=-2$ and $y_{2}=-1$.Comparing the values:$x$ values are $\{-3, -2\}$$y$ values are $\{-2, -1\}$Comparing each pair:$-3 In all cases,$x \le y$ is satisfied.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+5x+6=0$
$II.$ $y^{2}+3y+2=0$

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With respect to the roots of $x^{2}-x-2=0$,we can say that

The roots of the quadratic equation $2x^2 + 3x + 1 = 0$ are:

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Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}+5x+6=0$$II.$ $y^{2}+3y+2=0$