Indical equation of Frobenius Method

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So I'm getting ready for the exam by doing last years exam. It gives a DE which goes as following

$$(x^2-x)y''+(4x-2)y'+2y=0$$

This gives me:

$$\sum^\infty_{n=0}(n+r-1)(n+r)a_nx^{n+r}-\sum^\infty_{n=1}(n+r-1)(n+r-2)a_{n-1}x^{n+r}+\\4\sum^\infty_{n=0}(n+1)a_nx^{n+r}-2\sum^\infty_{n=1}(n+r-1)a_{n-1}x^{n+r}+2\sum^\infty_{n=0}a_nx^{n+r}$$

But, when finding the IE. Am I supposed to add the constants in front of the summation notation or am I supposed to leave it, like for some reason when I left it out I got the correct roots, but when I added it to the IE I got the wrong roots. The roots are supposed to be $(0,-1)$

Adding the constants:

$$((r-1)r+4r+2)a_0+\sum^\infty_{n=1}((n+r-1)(n+r)+4(n+r)+2)a_n-((n+r-2)(n+r-1)-2(n+r-1))a_{n-1}$$

Giving me the roots $(-1,-2)$

but when I don't add the constants i'll get

$$((r-1)r+r+1)a_0+\sum^\infty_{n=1}((n+r-1)(n+r)+4(n+r)+2)a_n-((n+r-2)(n+r-1)-2(n+r-1))a_{n-1}$$

which gives the roots $(0,-1)$

or is this method technically wrong and the first one should be the right one but i've managed to get lost in the algebra?

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Roots of indical equations are best obtained by letting $y=x^m$ in the ODE and then by setting the coefficient of the least power of $x$. Here the least power term you get is $$(-m^2+m-2m) x^{m-1}$$. Hence, the correct indical roots are $m=0,-1$.