We can find the roots of a quadratic equation with the quadratic formula
$x =\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$
For 1st equation
$x =\frac{-5 \pm \sqrt{25-24}}{12}$
$x =\frac{-5 \pm 1}{12}$
$x =-0.5,-0.33$
For 2nd equation
$15y^{2}+8y+1$=0
$15y^{2}+3y+5y+1$
y=$y=- \frac{1}{5} , -\frac{1}{3}$
(We can also find the roots by same method used for 1st equation.)
$y=- 0.2 , -0.33$
Here,
$- 0.2 \ , \ -0.33 \ \geq -0.5 \ , \ -0.33$
Hence relation between x and y = $y \geq x$
Hence,Option $(B) x\leq y .$ is the correct choice.