If $(a+ib{)}^{\frac{1}{4}}=2+3i,$ then $3b-2a=$

$\text{Given }(a+ib{)}^{\frac{1}{4}}=2+3i$

$\Rightarrow (a+ib{)}^{}={\left(2+3i\right)}^4$

$\Rightarrow a+ib= {\left[{\left(2+3i\right)}^2\right]}^2={\left[4+9i^2+12i\right]}^2$

$={\left[4-9+12i\right]}^2$

$={\left[12i-5\right]}^2$

$=144i^2+25-120i$

$=-119-120i$

Comparing with $a+ib$

$a=-119, b=-120$

$\text{Therefore }3b-2a=3(-120)-2(-119)=-122$