Divide both sides by -3, and replace [tex]x^2[/tex] with [tex]y[/tex]. Then
[tex]-3x^4+27x^2+1200=0\iff y^2-9y-400=0[/tex]
Factorize the quadratic in [tex]y[/tex] to get
[tex]y^2-9y-400=(y+16)(y-25)=0\implies y=-16,y=25[/tex]
which in turn means
[tex]x^2=-16,x^2=25[/tex]
But [tex]x^2\ge0[/tex] for all real [tex]x[/tex], so we can ignore the first solution. This leaves us with
[tex]x^2=25\implies x=\pm\sqrt{25}=\pm5[/tex]
If we allow for any complex solution, then we can continue with the solution we ignored:
[tex]x^2=-16\implies x=\pm\sqrt{-16}=\pm i\sqrt{16}=\pm4i[/tex]
