Range of $xyz\;,$ If $x+y+z=4$ and $x^2+y^2+z^2=6$ 
If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$

$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$
Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roots of cubic equation, Then
$$\displaystyle (t-x)(t-y)(t-z)=0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$
So we get $\displaystyle t^3-4t^2-5t-c=0$
Now let $f(t)=t^3-4t^2-5t-c\;,$ Then $f'(t)=3t^2-8t-5$
and $f''(t)=6t-8.$ Now for max. and Min.$f'(t)=0\Rightarrow 3t^2-8t-5=0$
So we get $\displaystyle t=\frac{8\pm \sqrt{64+60}}{2\cdot 3}=$
Now How can I solve it after that, Help required, Thanks

I thought it might be worth showing the Lagrange-multiplier method applied to this problem, largely for illustrating how the system of "Lagrange equations" can be handled, and for showing the interesting character of the solution.
With the constraints $ \ x^2 \ + \ y^2 \ + \ z^2 \ = \ 6 \ $ and $ \ x \ + \ y \ + \ z \ = \ 4 \ $ on the function $ \ f (x, \ y, \ z ) \ = \ xyz \ $ , the equations using two "multipliers" are
$$ yz \ = \ \lambda \cdot  2x \ + \ \mu \cdot 1 \ \ , \ \ xz \ = \ \lambda \cdot  2y \ + \ \mu \cdot 1 \ \ , \ \ xy \ = \ \lambda \cdot  2z \ + \ \mu \cdot 1 \ \ , $$
permitting us to write
$$ \mu \ = \ yz \ - \ 2 \lambda x \ = \ xz \ - \ 2 \lambda y \ = \ xy \ - \ 2 \lambda z \   \ . $$
Re-arranging the first implied equation produces
$$ \ (y \ - \ x) \ z \ + \ (y \ - \ x ) \ 2 \lambda \ = \ 0 \ \ \Rightarrow \ \ (y \ - \ x) \ ( z \ + \ 2 \lambda) \ = \ 0 \ \ ; $$
the other equated pairs of terms give us similar relations.
One solution then is to use $ \ x \ = \ y \ \ , \ \ x \ = \ z \ \ , $ and $ \ y \ = \ z \ $ in turn to obtain from the constraint equations
$$ x \ + \ x \ + \ z \ = \ 4 \ \ \Rightarrow \ \ x^2 \ + \ x^2 \ + \ ( 4 \ - \ 2x)^2 \ = \ 6x^2 \ - \ 16x \ + \ 16 \ = \ 6 $$
$$ 3x^2 \ - \ 8x \ + \ 5 \ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac{4 \ \pm \ 1}{3} \ = \ 1 \ \ , \ \ \frac{5}{3} \ \ , $$
as already described in the comments for the posted question, which the corrected cubic equation of OP would yield.  So we find three ordered triples, $ ( 1, \ 1, \ 2) \ , \ ( 1, \ 2, \ 1) \ , $ and $ \ ( 2, \ 1, \ 1) \ $ which give the same value of $ \ 2 \ $ for $ \ xyz \ $ and another three, 
$ ( \frac{5}{3}, \ \frac{5}{3}, \ \frac{2}{3}) \ , \ ( \frac{5}{3}, \ \frac{2}{3}, \ \frac{5}{3}) \ , $ and $ \ ( \frac{2}{3}, \ \frac{5}{3}, \ \frac{5}{3}) \ $ , for all of which $ \ xyz \ = \ \frac{50}{27} \ $ .
The alternative of using $ \ x \ \ne \ y \ , \ z \ = \ - 2 \lambda \ $ (and analogously for the other combinations of the three variables) produces the equation $ \ yz \ - \ 2 \cdot \left( -\frac{z}{2} \right) \cdot x \ = \ xz \ - \ 2 \cdot \left( -\frac{z}{2} \right) \cdot y \ $ $ \Rightarrow \ \ yz \ + \  xz \ = \ xz \ + \  yz \ $ , so no further information is gained.  We conclude that we have found the extremal values for the functions already and that our function has the (constrained) range $$ \ \frac{50}{27} \ \le \ f (x, \ y, \ z ) \ \le \ 2 \  \ . $$
The graph below shows the geometrical interpretation of a sphere intersected by a tilted plane, so that we are seeking extremal values of the function on a circle.  Since the function has symmetry about the line $ \ x \ = \ y \ = \ z \ $ , it may be expected that the maxima and minima number three each and are arranged symmetrically around the "constraint circle", the center of which lies at $ \ ( \frac{4}{3}, \ \frac{4}{3}, \ \frac{4}{3}) \ $ , which is connected to the appearance of $ \ \frac{4}{3} \ $ in the solution calculation using the quadratic formula in the original post.

The yellow "vertical" lines emerge at the positions of the maximal values for the function, while the red lines mark the locations of the minimal values.
