Find the maximum of $x$ satisfies the cubic equation Let $a,b,c,d$ be integers with $1\leq a,b,c,d\leq10$, and $x$ be real such that

$ax^3-bx^2-cx-d=0$. How to find the maximum of $x$ ?

Thanks in advance.
From Descartes' rule of signs, it is evident the cubic $f(x) = ax^3-bx^2-cx-d$ has exactly one positive root for all allowable $a, b, c, d$, so we just need to find its maximum value.  

Let $r$ be the positive root of $p(r)=r^3-10(r^2+r+1)$. It is clear $p(x)&gt; 0$ for $x&gt; r$, as $r$ is the only one positive root, where the sign turns from negative to positive.

Now if $x&gt;r$, we have $f(x)\ge p(x)&gt;p(r)=0$.  Thus $f(x)$ cannot have any root $&gt; r$.  

Numerically $r \approx 10.993$, but if an exact formula is needed, you need to use a method like Cardano's formula to get $r = \frac13 \left(10+\sqrt[3]{5 (317+3 \sqrt{1401})}+\sqrt[3]{5(317-3 \sqrt{1401})}\right)$.
discriminant of this equation is 
$c^2b^2-4(db^3+c^3a)+18abcd-27d^2b^2$ thus when the  discriminant will have max value that equation will have max value for $x$.or use vieta's formula