The polynomial is given to be:
[tex]3x^2+3x^3=6x^4[/tex]We can rewrite the polynomial to be:
[tex]6x^4-3x^3-3x^2=0[/tex]Factor the polynomial using the common factor of each term:
[tex]\begin{gathered} Factor\Rightarrow3x^2 \\ \therefore \\ \Rightarrow3x^2(2x^2-x-1)=0 \end{gathered}[/tex]Using the Zero Factor Principle, given to be:
[tex]\begin{gathered} If \\ ab=0 \\ \text{then} \\ a=0,b=0 \end{gathered}[/tex]Therefore, we have:
[tex]\begin{gathered} 3x^2=0,x^2=\frac{0}{3}=0,x=\pm\sqrt[]{0} \\ \therefore \\ x=0(mult.2) \end{gathered}[/tex]or
[tex]2x^2-x-1=0[/tex]Solving using the quadratic formula given to be:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{where} \\ a=2 \\ b=-1 \\ c=-1 \end{gathered}[/tex]Therefore, we have:
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt[]{(-1)^2-4\times2\times(-1)}_{}}{2\times2}=\frac{1\pm\sqrt[]{1+8}}{4}=\frac{1\pm\sqrt[]{9}}{4} \\ x=\frac{1+3}{4}=-\frac{4}{4}=1 \\ or \\ x=\frac{1-3}{4}=\frac{-2}{4}=-\frac{1}{2} \end{gathered}[/tex]Therefore, the solution to the equation is:
[tex]x=0(mult.2),1,-\frac{1}{2}[/tex]Comparing our answers with the provided options, we can check for equivalent answers. From the Third Option, we have:
[tex]\begin{gathered} x=\frac{3\pm\sqrt[]{81}}{12}=\frac{3\pm9}{12} \\ \therefore \\ x=\frac{3+9}{12}=\frac{12}{12}=1 \\ or \\ x=\frac{3-9}{12}=-\frac{6}{12}=-\frac{1}{2} \end{gathered}[/tex]This is equivalent to the answer we got.
Therefore, the correct option is the THIRD OPTION:
[tex]x=0(mult.2),\frac{3\pm\sqrt[]{81}}{12}[/tex]