Question 361774
If {{{3i}}} and {{{-2i}}} are solutions, then {{{(x-3i)}}} and {{{(x+2i)}}} are factors of the quadratic equation.
{{{(x-3i)(x+2i)=ax^2+bx+c}}}
{{{x^2+2ix-3ix-6i^2=ax^2+bx+c}}}
{{{x^2+(-3i+2i)x+6=ax^2+bx+c}}}
{{{x^2-ix+6=ax^2+bx+c}}}

Comparing
{{{a=1}}}
{{{b=-i}}}
{{{c=6}}}
However a,b, and c are supposed to be real. 
Therefore {{{3i}}} and {{{-2i}}} cannot be solutions where a,b, and c are real numbers.