<s>[INST] Solve $x^4+3x+20=0$ by Ferrari's method Comparing the equation 
$$x^4+3x+20=0$$
With the equation 
$$(x^2+\lambda)^2-(mx+n)^2=0$$
we get    
$m^2=2\lambda,$
$-2mn=3,$
$n^2=\lambda^2-20$   
Now, $4m^2n^2=9\Rightarrow 4(2\lambda)(\lambda^2-20)=9\Rightarrow 8\lambda^3-160\lambda-9=0$.   
How can I find easily the values of $\lambda$ from the above equation.
Please suggest me.
 [/INST] You can't find them easily, as the discriminant of the equation is standard form: 
$$x^3-20x+\frac 98=0$$
is $  4\cdot (-20)^3+ 27\cdot\Bigl(\dfrac 98\Bigr)<0$.
In such a case, we know the equation has $3$ real toots, but Cardano's formula requires using complex numbers since the square root is that of a negative number, then computing the cubic roots of a complex number.
A trigonometric method is more direct: set $x=A\cos t,\enspace A>0$. The equation becomes 
$$A^3\cos^3 t -20A\cos t+\frac 98=0.$$
We the choose $A>0$ so that $\dfrac{A^3}4=\dfrac{-20A}3$. Let's call $k$ this common ratio. We get:
$$k(4\cos^3t-3\cos t)=k\cos 3t=-\frac 98$$
whence
$$3t\equiv \pm\arccos\Bigl(-\frac 9{8k}\Bigr)\mod 2\pi\iff t\equiv\pm\frac 13\arccos\Bigl(-\frac 9{8k}\Bigr)\mod\frac{2\pi}3. $$
These $6$ solutions are equal in pairs, and there are $3$ roots in all, of course.
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