Question

What are the roots of the equation x^2-6x+58=0 in simplest a+bi form?

Answer 1

Explanation:

`x^2-6x+58=0\\ \\ \text{Since the discriminant, 36-232<0, there are no real roots, only two imaginary ones.}\\ \\ x=3\pm 7i\\ \\ (x+3+7i)(x+3-7i)`