Given that,
$\begin{aligned}
& x=\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 5}+\cdots \\
& =\frac{1}{5}+\frac{1 \cdot 3}{2 \times 1}\left(\frac{1}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 \times 2 \times 1}\left(\frac{1}{5}\right)^3+\cdots \\
& =\frac{1}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{1}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{1}{5}\right)^3+\cdots
\end{aligned}$
On adding 1 both the sides, we get
$1+x=1+\frac{1}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{1}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{1}{5}\right)^3+\cdots$
Now,
According to the binomial theorem for any index,
$\begin{aligned}
& \left(1-\frac{2}{5}\right)^{-1 / 2}=1+\frac{1}{5}+\frac{1 \cdot 3}{2 !}\left(\frac{1}{5}\right)^2+\frac{1 \cdot 3 \cdot 5}{3 !}\left(\frac{1}{5}\right)^3+\cdots \\
& 1+x=\left(1-\frac{2}{5}\right)^{-1 / 2} \Rightarrow 1+x=\left(\frac{3}{5}\right)^{-1 / 2} \\
& 1+x=\left(\frac{5}{3}\right)^{1 / 2} \\
& (1+x)^2=\frac{5}{3} \\
& 1+2 x+x^2=\frac{5}{3} \\
& 3+6 x+3 x^2=5 \\
& 3 x^2+6 x=2
\end{aligned}$