complex solve solutions of polynomial

Show that the solutions to the equation $2z^3-7z^2+10z-6=0$, given $1+i$ is a root, are $z=\frac{3}{2}$, $1+i$ and $1-i$.

All the coefficients are real, so each complex root must also have its conjugate.

$1+i$ and $1-i$ are roots.

$2z^3-7z^2+10z-6=0$

$(z-1+i)(z-1-i)=(z-1{)}^2-i^2=z^2-2z+2$

$0=(2z-3)(z^2-2z+2)$
$0=(2z-3)[(z-1{)}^2+1]$
$0=(2z-3)[(z-1{)}^2-(i{)}^2]$
$0=(2z-3)(z-1+i)(z-1-i)$