Solve this

Given

$2 x^{2}+3 i x+2=0$

$\Rightarrow 2 x^{2}+4 i x-i x+2=0$

$\Rightarrow 2 x(x+2 i)-i\left(x-\frac{2}{i}\right)=0$

$\Rightarrow 2 x(x+2 i)-i\left(x-\frac{2 \times i}{i \times i}\right)=0$

$\Rightarrow 2 x(x+2 i)-i\left(x-\frac{2 \times i}{-1}\right)=0$

$\Rightarrow 2 x(x+2 i)-i(x+2 i)=0$

$\Rightarrow(x+2 i)(2 x-i)=0$

$\Rightarrow x+2 i=0 \& 2 x-i=0$

$\Rightarrow \mathrm{x}=-2 \mathrm{i} \&^{x=\frac{i}{2}}$

Ans: $\mathrm{x}=-2 \mathrm{i}$ and $^{x}=\frac{i}{2}$