Answer
The solutions are $x=2\pm5i$
Work Step by Step
$x^{2}-4x+29=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=1$, $b=-4$ and $c=29$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(29)}}{2(1)}=\dfrac{4\pm\sqrt{16-116}}{2}=...$
$...=\dfrac{4\pm\sqrt{-100}}{2}=\dfrac{4\pm10i}{2}=2\pm5i$
