The solutions are:
(-2, 0, 3)
Explanation:Given the equations:
[tex]-2a^3+2a^2+12a=0[/tex]This can be written as:
[tex]-2a(a^2-a-6)=0[/tex]So that
[tex]\begin{gathered} -2a=0 \\ \Rightarrow a=0 \\ OR \\ a^2-a-6=0 \\ (a-3)(a+2)=0 \\ a-3=0 \\ \Rightarrow a=3 \\ \\ OR \\ a+2=0 \\ \Rightarrow a=-2 \end{gathered}[/tex]The solutions are:
(-2, 0, 3)
