Normal of line is parallel to line $x+90 y+2=0$
$m _{ N }=-\frac{1}{90} $
$ -\left(\frac{ dx }{ dy }\right)_{\left(x_1 y_1\right)}=-\frac{1}{90} \Rightarrow\left(\frac{ dy }{ dx }\right)_{\left(x_1 y_1\right)}=90$
Now,
$ \frac{d y}{d x}=270 x^4-540 x^3-210 x^2+360 x+210=90$
$ \Rightarrow x=1,2, \frac{-2}{3}, \frac{-1}{3}$
(4) normals