The number of points on the curve  $y=54x^5-135x^4-70x^3+180x^2+210x$ at which the normal lines are parallel to $x+90y+2=0$  is $l$ then  $\frac{l}{4}$ is 

Since, normal of line is parallel to line  $x+90y+2=0$  is  $m=-\frac{1}{90}$
 $\Rightarrow -{\left(\frac{dx}{dy}\right)}_{(x_1,y_1)}=-\frac{1}{90}\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(x_1{y}_1\right)}=90$
Now,  $\frac{dy}{dx}=270x^4-540x^3-210x^2+360x+210=90$
After solving we get, 
 $x=1,2,\frac{-2}{3},\frac{-1}{3}$
There are total 4 normals