Ta có : \(17x^2+805x+1800=0\)
=> \(\left(x\sqrt{17}\right)^2+2.x\sqrt{17}.\frac{805}{2\sqrt{17}}+\frac{648025}{68}-\frac{525625}{68}=0\)
=> \(\left(x\sqrt{17}+\frac{805}{2\sqrt{17}}\right)^2-\left(\sqrt{\frac{525625}{68}}\right)^2=0\)
=> \(\left(x\sqrt{17}+\frac{805}{2\sqrt{17}}-\sqrt{\frac{525625}{68}}\right)\left(x\sqrt{17}+\frac{805}{2\sqrt{17}}+\sqrt{\frac{525625}{68}}\right)=0\)
=> \(\left[{}\begin{matrix}x\sqrt{17}+\frac{805}{2\sqrt{17}}-\sqrt{\frac{525625}{68}}=0\\x\sqrt{17}+\frac{805}{2\sqrt{17}}+\sqrt{\frac{525625}{68}}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\sqrt{17}=-\frac{805}{2\sqrt{17}}+\sqrt{\frac{525625}{68}}\\x\sqrt{17}=-\frac{805}{2\sqrt{17}}-\sqrt{\frac{525625}{68}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{-\frac{805}{2\sqrt{17}}+\sqrt{\frac{525625}{68}}}{\sqrt{17}}=-\frac{40}{17}\\x=\frac{-\frac{805}{2\sqrt{17}}-\sqrt{\frac{525625}{68}}}{\sqrt{17}}=-45\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=-\frac{40}{17},x=-45\) .