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$