Question

does anyone know a function with two irrational solutions, and another with two complex solutions?

Answer 1

you can create
your self
let the roots be \[\sqrt{2},\sqrt{3},2\iota,-2\iota \]
then equation is
\[\left( x-\sqrt{2} \right)\left( x-\sqrt{3} \right)\left( x-2\iota \right)\left( x+2\sqrt{\iota } \right)=0\]
simplify it.

Answer 2

f(x) = x^2 + 1 would give you two complex roots.
(x + i) (x-i) =0
x= -i
x= i those two could be considered complex with 0 real values.

Answer 3

\[f(x)=x^2-5x+1\]
take any quadratic whose determinant >0 and is not a perfect square.