Question 882963
You may find it easy to factor {{{y^2+y-12=(y+4)(y-3)}}} .
If you replace {{{x^2}}} for {{{y}}} (and consequently {{{x^4}}} for {{{y^2}}} ),
you realize that {{{x^4+x^2-12=(x^2+4)(x^2-3)}}} and that
any solution to {{{x^4+x^2-12=0}}} is a solution to {{{(x^2+4)(x^2-3)=0}}} and vice versa.
The only real solutions to {{{(x^2+4)(x^2-3)=0}}} are
{{{highlight(x=sqrt(3))}}} and {{{highlight(x=-sqrt(3))}}} ,
which are the solutions to {{{x^2-3=0}}} .
{{{x^2+4=0}}} does not have any real solution.