A root of the equation $17x^2 + 17 x tan \left(2 tan^{-1}\frac{1}{5}-\frac{\pi}{4}\right)- 10 = 0, $ is

1

We have

$tan \left(2 tan^{-1}\frac{1}{5}-\frac{\pi}{4}\right)$

$= tan\begin{Bmatrix}tan^{-1}\frac{5}{12}-tan^{-1}1\end{Bmatrix}\left[∵ 2tan^{-1}= tan^{-1}\frac{2x}{1-x^2}\right]$

$= tan\begin{Bmatrix}tan^{-1}\left(\frac{\frac{5}{12}-1}{1+\frac{5}{12}}\right)\end{Bmatrix}=\frac{-7}{17}$

So, the given equation is

$17x^2 -7x - 10 = 0 ⇒ (x-1) (17x+10)= 0 ⇒x = 1, \frac{-10}{7}$