Step 1:
Given :$y=x^4-6x^3+13x^2-10x+5$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=$$4x^3-18x^2+26x-10$
The given points are $(0,5)$
Putting $x=0$
We get $\large\frac{dy}{dx}_{(0,5)}$$=-10$
Therefore slope of the tangent at $(0,5)$ is $-10$
Step 2:
Therefore equation of the tangent at $(0,5)$ is $(y-5)=-10(x-0)$
$\Rightarrow y-5=-10x$
$\Rightarrow 10x+y-5=0$
Step 3:
Therefore equation of the normal at $(0,5)$ is $(y-5)=\large\frac{-1}{10}$$(x-0)$
$\Rightarrow y-5=\large\frac{-1}{10}$$x$
$-10y+50=-x$
$\Rightarrow x-10y+50=0$