Question:
$x^{2}+2 x+5=0$
Solution:
We have:
$x^{2}+2 x+2=0$
$\Rightarrow x^{2}+2 x+1+1=0$
$\Rightarrow x^{2}+2 \times x \times 1+1^{2}-(i)^{2}=0$
$\Rightarrow(x+1)^{2}-(i)^{2}=0$
$\Rightarrow(x+1+i)(x+1-i)=0$
$\Rightarrow(x+1+i)=0$ or $(x+1-i)=0$
$\Rightarrow x=-1-i$ or $x=-1+i$
Hence, the roots of the equation are $-1+i$ and $-1-i$.
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