Question

determine the type number solution 3x^2+6x+4=0
a.1 real solution
b.one real and 1imaginary solution
c. 2 imaginary solution
d. 2 real solution

Answer 1

Sqrt[b^2 - 4ac]

Answer 2

that is 6^2-4*3*4=36-48=-12
sqrt of -12 is an imaginary number, so there are 2 imaginary solutions

Answer 3

by the way of there is an equation like x^n+x^(n-1)+...+1=0
it will always have n number of solutions if we allow complex numbers