A root of the equation 17x2+17xtan⁡2tan−1⁡15−π4−10=0, is

We have,tan⁡2tan−1⁡15−π4=tan⁡tan−1⁡512−tan−1⁡1∵2tan−1⁡x=tan−1⁡2x1−x2=tan⁡tan−1⁡512−11+512=−717So, the given equation is17x2−7x−10=0⇒(x−1)(17x+10)=0⇒x=1,−107