Question:
$x^{2}-4 x+7=0$
Solution:
We have:
$x^{2}-4 x+7=0$
$\Rightarrow x^{2}-4 x+4+3=0$
$\Rightarrow x^{2}-2 \times x \times 2+2^{2}-(\sqrt{3} i)^{2}=0$
$\Rightarrow(x-2)^{2}-(\sqrt{3} i)^{2}=0$
$\Rightarrow(x-2+\sqrt{3} i)(x-2-\sqrt{3} i)=0$
$\Rightarrow(x-2+\sqrt{3} i)=0$ or, $(x-2-\sqrt{3} i)=0$
$\Rightarrow x=2-\sqrt{3} i \quad$ or, $\quad x=2+\sqrt{3} i$
Hence, the roots of the equation are $2 \pm i \sqrt{3}$.
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