Applying Descartes's method, I determined the interim equation as $y^4-3y^2-28=0$. Then I went on to treat this as a product of $(y^2+ky+m)(y^2-ky+n)$. Comparing coefficients of $y^2$, $y$ and constant term, I obtained $m+n-k^2=-3$, $k(n-m)=0$ and $mn=-28$. I am unable to proceed any further.
On the other hand, if I express $y^4-3y^2-28=0$ as $(y^2+7)(y^2-4)$, I get the roots as $-1$, $3$, $1+i\sqrt{7}$ and $1-i\sqrt{7}$.
Apparently, $1+i\sqrt{7}$ and $1-i\sqrt{7}$ are the correct roots. The equation seems to have two other complex roots which I am unable to find.
Appreciate any help. Where am I going wrong?
$$ y^4-3y^2-28 = (y^2-7)(y^2 + 4) $$ And $x=y+1$, so roots are $1\pm\sqrt7$, $1\pm 2i$.