First of all, you probably have a mistake in what you did: if $L(p(x))=p(0)x-p(1)$ and $p(x)=ax^2+bx+c$, then $L(p(x))=cx-(a+b+c)$.
For the kernel you can proceed as xavierm02 did in the comments. A polynomial $p$ is in the kernel of $L$ if $L(p)=0$. Thus assuming that $p(x)=ax^2+bx+c$ you have that $c=0$ and $a+b+c=0$ by comparing coefficients. Now you can solve this system of linear equations to get a basis of the kernel.
For the range it is obvious that there is no polynomial with quadratic coefficient in the range. For the (at most) linear one's it should be pretty easy to find preimages either you do that generally, or you give a preimage of $x$ and $1$ and then argue with linearity, or you use the rank-nullity theorem and conclude it is two-dimensional so it has to be the whole space of (at most) linear polynomials.