Find the equation

finding the slope of the tangent by differentiating the curve

$\frac{d y}{d x}=3 x^{2}+2$

$\mathrm{m}($ tangent $)=3 x^{2}+2$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$\mathrm{m}($ normal $)=\frac{-1}{3 \mathrm{x}^{2}+2}$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

now comparing the slope of normal with the given equation

$\mathrm{m}($ normal $)=-\frac{1}{14}$

$-\frac{1}{14}=-\frac{1}{3 x^{2}+2}$

$x=2$ or $-2$

hence the corresponding value of $y$ is 18 or $-6$

so, equations of normal are

$y-18=-\frac{1}{14}(x-2)$

Or

$y+6=-\frac{1}{14}(x+2)$