Question 131774
One way is to use the quadratic formula and calculate the discriminant.


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


The discriminant is the {{{b^2-4*a*c}}} part under the radical.


If {{{b^2-4*a*c>0}}}, there are two real and unequal roots.


If {{{b^2-4*a*c=0}}}, there are two real and equal roots.  (Also known as one real root with a multiplicity of two.  There is never just one real solution to a quadratic, there are always two, it's just that sometimes the two are equal.  Google "Fundamental Theorem of Algebra" for more information.)


If {{{b^2-4*a*c<0}}}, there is a conjugate pair of complex roots.


So how does that apply to your equation?


{{{x^2 + 3 = 0 }}}.  There is no 'x' term in this equation.  That means the b coefficent in the quadratic formula is 0.  You could actually write your equation as:


{{{x^2 + 0x + 3=0}}}


Let's look at the discriminant:


{{{b^2-4*a*c}}}
{{{0^2-4*1*3=-12<0}}}


Since the discriminant is < 0, there is a conjugate pair of complex roots of the form {{{a+bi}}} and {{{a-bi}}} where {{{i^2=-1}}}.  In the case of your problem, the a in {{{a+bi}}} will turn out to be 0, and {{{b=sqrt(3)}}}.