With
[tex]y=\displaystyle\sum_{n\ge0}a_nx^{n+r}[/tex]
[tex]y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}[/tex]
[tex]y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}[/tex]
the singular ODE can be written as
[tex]\displaystyle\sum_{n\ge0}\bigg[2(n+r)(n+r-1)+3(n+r)-1\bigg]a_nx^{n+r}+\sum_{n\ge0}\bigg[-2(n+r)(n+r-1)-3(n+r)\bigg]a_nx^{n+r-1}=0[/tex]
The first term of the second series admits the indicial equation. When [tex]n=0[/tex], we have
[tex](-2r(r-1)-3r)a_0x^{r-1}=0\iff-2r^2-r=0\iffr^2+\dfrac12r=0[/tex]
Factoring reveals two distinct roots at [tex]-r(2r+1)=0\implies r_1=0,r_2=-\dfrac12[/tex] (in your case, swap [tex]r_1[/tex] and [tex]r_2[/tex] before submitting).
Next, shift the index of the first sum so that it starts at [tex]n=1[/tex] by replacing [tex]n\mapsto n-1[/tex], then consolidate the sums to get
[tex]\displaystyle\sum_{n\ge1}\bigg[2(n+r-1)(n+r-2)+3(n+r-1)-1-2(n+r)(n+r-1)-3(n+r)\bigg]a_nx^{n+r-1}[/tex]
[tex]=\displaystyle x^r\sum_{n\ge1}\bigg[-4n-4r\bigg]a_nx^{n-1}[/tex]
Setting [tex]r=-\dfrac12[/tex], we then have this as
[tex]=x^{-1/2}\left(-2a_1-6a_2x-10a_3x^2-14a_4x^3-18a_5x^4+\cdots\right)[/tex]
[tex]=x^{-1/2}\displaystyle\sum_{n\ge1}(2-4n)a_nx^{n-1}[/tex]
However I don't see the connection to the given answer... It seems some information is missing, specifically about how the coefficients [tex]a_n[/tex] are related.
