Step 1:

$2xy+3x+4y+1=0$

The above equation is divided by $2$ we get,

$xy+\large\frac{3}{2}$$x+2y+\large\frac{1}{2}=$$0$

Step 2:

Now we take the LHS of the above equation

LHS=$x(y+\large\frac{3}{2})$$+2(y+\large\frac{3}{2})$$-3+\large\frac{1}{2}$

$\qquad=(x+2)(y+\large\frac{3}{2})-\large\frac{5}{2}$

$\Rightarrow (x+2)(y+\large\frac{3}{2})-\large\frac{5}{2}$$=0$

$(x+2)(y+\large\frac{3}{2})=\large\frac{5}{2}$

Step 3:

Since the equation is in standard form,the equations of the asymptotes are $x+2=0,y+\large\frac{3}{2}$$=0$