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

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Answer 1 · 2015-03-02T14:45:31

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 $sqrt(-1)=i$ .

How do I find all real and complex zeros of x^2+3x+12x2+3x+12x^2+3x+12?