Answer
$x=2-i
\text{ OR }
x=2+i$
Work Step by Step
The given equation, $x^2+5=4x,$ is equivalent to
\begin{align*}
x^2-4x+5&=0
.\end{align*}
Using $ax^2+bx+c=0,$ the equation above has $a=
1
,$ $b=
-4
,$ and $c=
5
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}\require{cancel}x&=
\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(5)}}{2(1)}
\\\\&=
\dfrac{4\pm\sqrt{16-20}}{2}
\\\\&=
\dfrac{4\pm\sqrt{-4}}{2}
\\\\&=
\dfrac{4\pm\sqrt{-1}\cdot\sqrt{4}}{2}
\\\\&=
\dfrac{4\pm\sqrt{-1}\cdot2}{2}
\\\\&=
\dfrac{4\pm i\cdot2}{2}
&\text{ (use $i=\sqrt{-1}$)}
\\\\&=
\dfrac{4\pm 2i}{2}
\\\\&=
\dfrac{\cancel4^2\pm \cancel2^1i}{\cancel2^1}
&\text{ (divide by $2$)}
\\\\&=
2\pm i
.\end{align*}
The solutions are $
x=2-i
\text{ and }
x=2+i
.$