An algebra problem by Phudish Prateepamornkul

$\begin{aligned} \frac 3{x-3} + \frac 5{x-5} + \frac {17}{x-17} + \frac {19}{x-19} & = x^2 - 11x - 4 \\ \frac {x-(x-3)}{x-3} + \frac {x-(x-5)}{x-5} + \frac {x-(x-17)}{x-17} + \frac {x-(x-19)}{x-19} & = x^2 - 11x - 4 \\ \frac x{x-3} + \frac x{x-5} + \frac x{x-17} + \frac x{x-19} - 4 & = x^2 - 11x - 4 \\ {\color{#D61F06}x}\left(\frac 1{x-3} + \frac 1{x-5} + \frac 1{x-17} + \frac 1{x-19}\right) & = {\color{#D61F06}x}(x - 11) & \small \color{#D61F06} \implies x = 0 \text{ is a root.} \\ \frac 1{x-3} + \frac 1{x-5} + \frac 1{x-17} + \frac 1{x-19} & = x - 11 & \small \color{#3D99F6} \text{Let }y = x-11 \\ {\color{#3D99F6}\frac 1{y+8}} + \frac 1{y+6} + \frac 1{y-6} + {\color{#3D99F6}\frac 1{y-8}} & = y \\ {\color{#3D99F6}\frac {2{\color{#D61F06}y}}{y^2-64}} + \frac {2{\color{#D61F06}y}}{y^2-36} & = \color{#D61F06}y & \small \color{#D61F06} \implies y = 0 \implies x = 11 \text{ is a root.} \\ \frac {4y^2 - 200}{(y^2-64)(y^2-36)} & = 1 \\ 4y^2 - 200 & = y^4 - 100y^2 + 2304 \\ y^4 - 104y^2 + 2504 & = 0 \\ \implies y^2 & = 52 \pm \sqrt {200} \\ y & = \pm \sqrt{52 \pm \sqrt {200}} \\ \implies x & = 11 \pm \sqrt{52 \pm \sqrt {200}} & \small \color{#D61F06} \text{4 more roots.} \end{aligned}$

The largest real root is $x=11+\sqrt{52 + \sqrt {200}}$ , $\implies a+b+c = 11+52+200 = \boxed{263}$ .