Note: This answer is based on the relation between the roots of a quadratic equation with real coefficients, in the case when both are complex numbers.
I used below $z_1,z_2$, instead of $z,z^{\ast}$.
If one of the roots of a quadratic equation is the complex number $z_1=x_1+iy_1$ the other root $z_2=x_2+iy_2$ is the conjugate of $z_1$, i.e $z_2=x_1-iy_1=z_1^{\ast}$, as it can be seen, among other ways, by the resolvent formula
$az^2 + bz + c = 0 \iff z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\qquad(a \ne 0).$
Thus, if
$z_1=\frac{-b+\sqrt{b^2-4ac}}{2a}=x_1+iy_1,$
then
$z_2=\frac{-b-\sqrt{b^2-4ac}}{2a}=x_1-iy_1=z_1^{\ast}$
and vice-versa.
Since $z_1,z_2$ are roots, both satisfy the quadratic equation, by definition.
Added in view of the comments below.
- If $z_2=z_1$, then $z_2=x_2=x_1$ and $y_1=y_2=0$.
- If $a=0$, then the equation reduces to $bz_1+c=0$, which has a single root $z_1=x_1$, and $y_1=0$.