Given the equation:
![6x=2x^3+11x^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/lk9kflr2kguqi9n7e1o7oieyh74tepco1y.png)
Let's solve the equation for x.
Rearrange the equation:
![2x^3+11x^2=6x](https://img.qammunity.org/2023/formulas/mathematics/high-school/bwybwd2ay9jos24l8hkqwbvnzhhsz5f0oo.png)
Subtract 6x from both sides:
![\begin{gathered} 2x^3+11x^2-6x=6x-6x \\ \\ 2x^3+11x^2-6x=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/qg9cmczk5glhroc8fq1lq6bveb94dkjmzn.png)
Now, let's factor the left side of the equation.
Factor out x from all terms:
![x(2x^2+11x-6)=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/tp42x5zklgvx9btn2qmgpcxgmeqlef4ys2.png)
The next step is to factor by grouping.
We have:
![\begin{gathered} x(2x^2-1x+12x-6)=0 \\ \\ x((2x^2-1x)(12x-6))=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/3whserr67m6rdyadseftc835ppwrer2o4p.png)
factor out x from the first group and factor out 6 from the second group:
![\begin{gathered} x(x(2x-1)+6(2x-1))=0 \\ \\ x((x+6)(2x-1))=0 \\ \\ x(x+6)(2x-1)=0 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2x9shzuomhqhy2d1jomgzjtwyefvkoe0si.png)
Thus, we now have the solutions:
![\begin{gathered} x=0 \\ x+6=0 \\ 2x-1=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/x0oo0oegqj8vbiqu816bvsk1rj9v5jgwkm.png)
Solve each factor for x.
We have:
![\begin{gathered} x=0 \\ \\ \\ x+6=0 \\ \text{ Subtract 6 from both sides:} \\ x+6-6=0-6 \\ x=-6 \\ \\ \\ 2x-1=0 \\ Add\text{ 1 to both sides:} \\ \text{ 2x - 1 + 1 = 0+1} \\ 2x=1 \\ \text{ Divide both sides by 2:} \\ (2x)/(2)=(1)/(2) \\ x=0.5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/l08ucp8q6wg7er6jx05o3f1bwx9pj4n02w.png)
Therefore, the solutions are:
x = -6, 0, 0.5
ANSWER:
x = -6, 0, 0.5