# Solve the following

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 $-1-i$.