Question

one question

Answer 1

For quadratic equations wrote as \[ax^2+bx+c=0\]
the discriminant is\[ b^2-4ac\]
If
\[ b^2-4ac>0\] your function has two real valued roots(or zeros, or solutions, whatever you called 'em)
\[ b^2-4ac=0\] then you have two EQUAL real valued roots, x1=x2
\[ b^2-4ac<0\] Complex stuff (You'll have two complex solutions)

Now, for \[x^2+6x-3=0\] a=1, b=6 and c=-3
\[6^2-4\times1\times-3=6^2+12=36+12=48>0\]
48 is greater than 0, so you have two real and different solutions.
(Which, by the way, are\[x_{1}=-3-2\sqrt {3} \ \ and \ \ x_{2}=2\sqrt {3} -3\])
Salut