Question:
$x^{2}+2 x+5=0$
Solution:
Given: $x^{2}+2 x+5=0$
$x^{2}+2 x+5=0$
$\Rightarrow x^{2}+2 x+1+4=0$
$\Rightarrow(x+1)^{2}-(2 i)^{2}=0 \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$\Rightarrow(x+1+2 i)(x+1-2 i)=0 \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$
$\Rightarrow(x+1+2 i)=0$ or, $(x+1-2 i)=0$
$\Rightarrow x=-(1+2 i)$ or, $x=-1+2 i$
Hence, the roots of the equation are $-1+2 i$ and $-1-2 i$.
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