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(D) For equation $I: x^{2}-2x-15=0$We factorize the quadratic equation: $x^{2}-5x+3x-15=0$$x(x-5)+3(x-5)=0$$(x-5)(x+3)=0$Thus,the roots are $x = 5$ an

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

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(D) For equation $I: x^{2}-2x-15=0$We factorize the quadratic equation: $x^{2}-5x+3x-15=0$$x(x-5)+3(x-5)=0$$(x-5)(x+3)=0$Thus,the roots are $x = 5$ and $x = -3$.For equation $II: y^{2}+5y+6=0$We factorize the quadratic equation: $y^{2}+3y+2y+6=0$$y(y+3)+2(y+3)=0$$(y+3)(y+2)=0$Thus,the roots are $y = -3$ and $y = -2$.Comparing the values:If $x = 5$,then $x > y$ (since $5 > -3$ and $5 > -2$).If $x = -3$,then $x = y$ (when $y = -3$) and $x Since the relationship varies depending on the chosen roots,the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I. x^{2}-2x-15=0$
$II. y^{2}+5y+6=0$

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Solve the given two equations and select the correct answer from the given options.$I. x^{2}-2x-15=0$$II. y^{2}+5y+6=0$