The root of the equation $2(1+\mathrm{i}){\mathrm{x}}^2-4(2-\mathrm{i})\mathrm{x}-5-3\mathrm{i}=0,\mathrm{where} \mathrm{i}=\sqrt{-1},$which has greater modulus, is

The given equation is 
$2(1+\mathrm{i}){\mathrm{x}}^2-4(2-\mathrm{i})\mathrm{x}-5-3\mathrm{i}=0$
$$\begin{gathered} \Rightarrow \mathrm{x}=\frac{4(2-\mathrm{i})\pm \sqrt{16(2-\mathrm{i}{)}^2+8(1+\mathrm{i}\left)\right(5+3\mathrm{i})}}{4(1+\mathrm{i})} \\ =\frac{4(2-\mathrm{i})\pm \sqrt{48-64\mathrm{i}+40-24+64\mathrm{i}}}{4(1+\mathrm{i})} \\ =\frac{4(2-\mathrm{i})\pm \sqrt{64}}{4(1+\mathrm{i})}=\frac{8-4\mathrm{i}\pm 8}{4(1+\mathrm{i})} \\ =-\frac{\mathrm{i}}{1+\mathrm{i}} \mathrm{or} \frac{4-\mathrm{i}}{1+\mathrm{i}}=\frac{-1-\mathrm{i}}{2} \mathrm{or} \frac{3-5\mathrm{i}}{2} \end{gathered}$$
Now,$\left|\frac{-1-i}{2}\right|=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{1}{2}}$
$\text{ and }\left|\frac{3-5\mathrm{i}}{2}\right|=\sqrt{\frac{9}{4}+\frac{25}{4}}=\sqrt{\frac{17}{2}}$
Also, $\sqrt{\frac{17}{2}}>\sqrt{\frac{1}{2}}$
Hence, required root is $\frac{3-5\mathrm{i}}{2}$