Solve this

Given:

$x^{2}+13=4 x$

$\Rightarrow x^{2}-4 x+13=0$

Solution of a general quadratic equation $a x^{2}+b x+c=0$ is given by:

$x=\frac{-b \pm \sqrt{\left(b^{2}-4 a c\right)}}{2 a}$

$\Rightarrow x=\frac{-(-4) \pm \sqrt{(-4)^{2}-(4 \times 1 \times 13)}}{2 \times 1}$

$\Rightarrow x=\frac{4 \pm \sqrt{16-52}}{2}$

$\Rightarrow x=\frac{4 \pm \sqrt{-36}}{2}$

$\Rightarrow x=\frac{4 \pm 6 i}{2}$

$\Rightarrow x=\frac{4}{2} \pm \frac{6}{2} i$

$\Rightarrow x=2 \pm 3 i$

Ans: $x=2+3 i \& x=2-3 i$