Let the speed of the train be xkmph

The time taken by the train to travel $180 \mathrm{~km}$ is $\frac{180}{\mathrm{x}} \mathrm{h}$

The increased speed is $\mathrm{x}+9$

The time taken is $\frac{180}{\mathrm{x}+9}$

According to the question,

The time taken is $\frac{180}{\mathrm{x}}-1$
$\Longrightarrow \frac{180}{\mathrm{x}}-1=\frac{180}{\mathrm{x}+9}$
$\Longrightarrow \frac{180-\mathrm{x}}{\mathrm{x}}=\frac{180}{\mathrm{x}+9}$
$\Longrightarrow 180 \mathrm{x}-\mathrm{x}^{2}+1680-9 \mathrm{x}=180 \mathrm{x}$
$\Longrightarrow \mathrm{x}^{2}+9 \mathrm{x}-1680=0$
$\Longrightarrow \mathrm{x}^{2}+45 \mathrm{x}-36 \mathrm{x}-1680=0$
$\Longrightarrow \mathrm{x}(\mathrm{x}+45)-36(\mathrm{x}+45)=0$
$\mathrm{x}=36,-45$

Speed cannot be negative (In this case,)

So The speed of the train is $36 \mathrm{kmph}$