Question

How do I find all real and complex zeros of `x^2+3x+12`?

Answer 1

First of all, let's find the discriminant:

`Delta=b^2-4ac=9-48=-39`.

Since `Delta<0`, the equation has two complex conjugate solution:

`x=(-b+-sqrtDelta)/(2a)=(-3+-sqrt(-39))/2=(-3+-sqrt39sqrt(-1))/2=`

`=-3/2+-isqrt39/2`.

Remember that `sqrt(-1)=i`.