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

 

We have $\tan \left(2{\tan }^{-1} (1/5)\right)$

$=\frac{2\times \frac{1}{5}}{1-\frac{1}{25}}=\frac{2}{5}\times \frac{25}{24}=\frac{5}{12}$

So that the given equation can be written as

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}17x^2-17x\frac{1-(5/12)}{1+(5/12)}-10=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}17x^2-7x-10=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(x-1)(17x+10)=0\end{array}$

$\therefore x=1$ is a root of the given equation.