Answer
See explanation
Work Step by Step
It is given a quadratic formula $ax^2+bx+c=0$ where $a,b,$ and $c$ are real numbers.
Suppose that the equation has complex roots $x_1$ and $x_2$.
It gives $D=b^2-4ac<0$ and so $-D>0$.
Using the quadratic formula,
$x=\frac{-b\pm \sqrt{D}}{2a}$
$x=\frac{-b\pm \sqrt{-1}\sqrt{-D}}{2a}$
$x=\frac{-b\pm \sqrt{-D}i}{2a}$
We know that both $p=-\frac{b}{2a}$ and $q=\frac{\sqrt{-D}}{2a}$ are real.
Then, $x_1=p+qi$ and $x_2=p-qi$.
Therefore, the roots $x_1$ and $x_2$ are complex conjugates of each other.
