Answer
$x\in \{-\sqrt 5i, \sqrt 5i\}$, each with a multiplicity of $2$
Work Step by Step
$Q(x)=x^4+10x^2+25$
The polynomial is the perfect square:
$Q(x)=x^4+10x^2+25=(x^2+5)^2=(x^2+5)(x^2+5)=(x-\sqrt 5i)(x+\sqrt 5i)(x-\sqrt 5i)(x+\sqrt 5i)$,
thus
$x^4+10x^2+25=(x-\sqrt 5i)^2(x+\sqrt 5i)^2$
Its zeros are:
$x\in \{-\sqrt 5i, \sqrt 5i\}$, each with a multiplicity of $2$