Solve the following

We have:

$4 x^{2}-12 x+25=0$

$\Rightarrow 4 x^{2}-12 x+9+16=0$

$\Rightarrow(2 x)^{2}+3^{2}-2 \times 2 x \times 3-(4 i)^{2}=0$

$\Rightarrow(2 x-3)^{2}-(4 i)^{2}=0$

$\Rightarrow(2 x-3+4 i)(2 x-3-4 i)=0 \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$

$\Rightarrow(2 x-3+4 i)=0$ or, $(2 x-3-4 i)=0$

$\Rightarrow 2 x=3-4 i$ or, $2 x=3+4 i$

$\Rightarrow x=\frac{3}{2}-2 i$ or, $x=\frac{3}{2}+2 i$

Hence, the roots of the equation are $\frac{3}{2}-2 i$ and $\frac{3}{2}+2 i$.