This equation has the next form:
[tex]ax^2+bx+c=0[/tex]To find if the equation has two complex solutions we have to check if the discriminant is negative, as follows:
[tex]\begin{gathered} b^2-4ac \\ (-5)^2-4\cdot3\cdot8=25-96=-71<0 \end{gathered}[/tex]Then, the first case has two complex solutions.
In the second case,
[tex]\begin{gathered} 2x^2=6x-5 \\ 2x^2-6x+5=0 \end{gathered}[/tex]The discriminant in this case is:
[tex](-6)^2-4\cdot2\cdot5=36-40=-4<0[/tex]Then, the second case has two complex solutions.
In the third case,
[tex]\begin{gathered} 12x=9x^2+4 \\ -9x^2+12x-4=0 \end{gathered}[/tex]The discriminant in this case is:
[tex]12^2-4\cdot(-9)\cdot(-4)=144-144=0[/tex]Then, the third case has two real solutions.
In the fourth case,
[tex]\begin{gathered} -x^2-10x=34 \\ -x^2-10x-34=0 \end{gathered}[/tex]The discriminant in this case is:
[tex](-10)^2-4\cdot(-1)\cdot(-34)=100-136=-36<0[/tex]Then, the fourth case has two complex solutions.