Q.Solve the quadratic equation $x^{2}-4x+13=0$ in the complex number system.

The discriminant is negative, so the roots are complex conjugates found via $\sqrt{-k}=\sqrt{k}\,i$.

Apply the quadratic formula to $x^{2}-4x+13=0$ (with $a=1,b=-4,c=13$):

Since $\sqrt{-36}=6i$,

The two roots are the complex conjugates $2+3i$ and $2-3i$. …

Completing the square: $x^{2}-4x+13=(x-2)^{2}+9=0\Rightarrow(x-2)^{2}=-9\Rightar …

Concluding 'no solution' because the discriminant is negative — in $\mathbb{C}$ it always has solutions. Also, dividing only one term of $ …