Applying Ferrari's method to a quartic

Question

I was interested in early solutions to quartics and cubics, so I read about Ferrari's method for solving the quartic. The book I read gives no specific example, but I was working through the equation

$$x^{4}+6x^{3}=6x^{2}+30x+11$$

or

$$x^{4}+6x^{3}-6x^{2}-30x-11=0$$

According to his method, the first thing we need to do is get rid of the $x^{3}$ by substituting, is this the correct substitution? Where do I go after this?$$x=v-\frac{6}{4}$$

So,

$$(v-\frac{6}{4})^{4}+6(v-\frac{6}{4})^{3}-6(v-\frac{6}{4})^{2}-30(v-\frac{6}{4})-11=0$$

Answer 1

...is this the correct substitution?

Yes

Where do I go after this?

Substitute: \begin{align*} x^4 + 6x^3 - 6x^2 - 30x - 11 &= 0\\ \left(v - \frac64\right)^4 + 6\left(v - \frac64\right)^3 - 6\left(v - \frac64\right)^2 - 30\left(v - \frac64\right) - 11 &= 0\\ &\vdots \end{align*}