172. $5 \cdot 5^{2x}-6 \cdot 5^{x}+1=0$
Solution. Let $5^{x}=y;$ then we get $5 y^{2}-6 y+1=0$. Here $a=5, \quad b=-6, \quad c=1 ; \quad D=b^{2}-4 a c=36-4 \cdot 5 \cdot 1=36-20=16$, $\sqrt{D}=4$. Therefore, $y_{1}=\frac{6-4}{10}=\frac{2}{10}=\frac{1}{5}, y_{2}=\frac{6+4}{10}=\frac{10}{10}=1$.

Since $y_{1}=1 / 5, y_{2}=1$, we have $5^{x}=1 / 5,5^{x}=1 ;$ then $5^{x}=1 / 5$, $5^{x}=5^{-1}$, i.e., $x=-1 ; 5^{x}=1,5^{x}=5^{0}$, i.e., $x=0$. Thus, we get the answer: $x=-1, x=0$.