Step 1:

The probability density function of a random variable $X$

$f(x)=Ce^{x^2-3x},-\infty < x < \infty$

Step 2:

$f(x)=Ce^{(-x^2-3x+\Large\frac{9}{4}-\frac{9}{4})}$

$\quad\quad=Ce^(-x^2-3x+\large\frac{9}{4})$$+\large\frac{9}{4}$

$\quad\quad=Ce^{\Large\frac{9}{4}}.e^{-(x-\Large\frac{3}{2})^2}$

$\quad\quad=Ce^{\Large\frac{9}{4}}.e^{-\large\frac{1}{2}\big(\Large\frac{x-\large\frac{3}{2}}{\Large\frac{1}{\sqrt 2}}\big)^2}$

Step 3:

Comparing with the probability density function $f(x)=\large\frac{-1}{\sigma \sqrt{2\pi}}$$e^{-\Large\frac{1}{2}(\Large\frac{x-\mu}{\sigma})^2}$

We can see that $\mu=\large\frac{3}{2}$$,\sigma=\large\frac{1}{\sqrt 2}$

$\large\frac{1}{\sigma \sqrt{2\pi}}=\frac{1}{\Large\frac{1}{\sqrt 2}.\sqrt 2\pi}$

$\qquad\;\;=\large\frac{1}{\sqrt{\pi}}$

$\qquad\;\;=Ce^{\Large\frac{9}{4}}$

$\large\frac{1}{\sqrt\pi}$$=Ce^{\Large\frac{9}{4}}$

$C=\large\frac{e^{-2.25}}{\sqrt{\pi}}$