Use your brains

I used Newton-Raphson.

First I tried to find the regions where the function changes sign and thus has a root.

$f(-1)=-1000-1254+496+191 < 0$

$f(0)= 191 > 0$

$f(1)=1000-1254-496+191 < 0$

$f(2)=8000-5016-992+191 > 0$

so we have one root in the region $[-1,0]$ , one root in the region $[0,1]$ and the last one in region $[1,2]$ . We want this last root.

Newton-Rhapson says that the next approximation for the root is $x_{n+1}$ according to the formula:

$x_{n+1} = x_n -\dfrac{f(x_n)}{f'(x_n)} =x_n- \dfrac{1000 x_n^3-1254x_n^2-496x_n+191}{3000x_n^2-2508x_n-496}$

Starting with $x_1 = 1.5$ , the middle of the interval $[1,2]$ , we get the following very fast converging series:

$x_1 = 1.5$

$x_2= 1.5 - \dfrac{f(1.5)}{f'(1.5)} = 1.5 - \dfrac{0.5}{2492} \approx 1.499799$

$x_3 = 1.499799 - \dfrac{f(1.499799)}{f'(1.499799)} = 1.499799 - \dfrac{0.0001306669}{2490.69755} \approx 1.499799$

So the solution is rounded to 4 decimals: $\boxed{1.4998}$ .

Let's make the substitution $x=\dfrac{y+1254}{3000}$ to delete the quadratic term:

$1000\left(\dfrac{y+1254}{3000}\right)^3-1254\left(\dfrac{y+1254}{3000}\right)^2-496\left(\dfrac{y+1254}{3000}\right)+191=0$

Expand, clear denominators and simplify:

$y^3-9181548y-4384726128=0$

Now, try to match that equation with the identity $(u+v)^3-3uv(u+v)-(u^3+v^3)=0$ , an equation system will be formed:

$\quad \begin{aligned} y=u+v &\quad ...(1) \\ 3uv=9181548 &\quad ...(2) \\ u^3+v^3=4384726128 &\quad ...(3) \end{aligned}$

Divide $(2)$ by $3$ and cube both sides:

$u^3v^3=28667113297167468096$

Now, make an equation in $z$ with roots $u^3$ and $v^3$ :

$(z-u^3)(z-v^3)=0 \Longrightarrow z^2-(u^3+v^3)z+u^3v^3=0$

$z^2-4384726128z+28667113297167468096=0$

Using the quadratic formula we get two values of $z$ :

$z=2192363064 \pm 6000i\sqrt{662796041466}$

Hence:

$u=\sqrt[3]{2192363064+6000i\sqrt{662796041466}}$

$v=\sqrt[3]{2192363064-6000i\sqrt{662796041466}}$

Using complex numbers let's take the principal cube roots for $u$ and $v$ :

$u=\sqrt[3]{|u|} cis \left(\dfrac{\arg u}{3}\right) \approx 1622.698958+653.730901i$

$v=\sqrt[3]{|v|} cis \left(\dfrac{\arg v}{3}\right) \approx 1622.698958-653.730901i$

From the equation system, there are three possible values for $(u,v)$ . They are: $(u,v)$ , $(uw,vw^2)$ and $(uw^2,vw)$ , where $w$ is any primitive 3rd root of unity. Can you check the other values that don't work?

So, we have three solutions for $y$ :

$y_1=u+v \approx 3245.397916$

$y_2=uw+vw^2 \approx -2754.994093$

$y_3=uw^2+vw \approx -490.4038231$

Hence:

$x_1=\dfrac{y_1+1254}{3000} \approx 1.4997$

$x_2=\dfrac{y_2+1254}{3000} \approx -0.5003$

$x_3=\dfrac{y_3+1254}{3000} \approx 0.2545$

From these solutions, the largest is $\boxed{1.4997}$ .