Question:
$4 x^{2}-12 x+25=0$
Solution:
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 $\frac{3}{2}+2 i$.
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