Question 546238
For any {{{x}}} that is a real number, {{{x^2>=0}}}, so {{{x^2+6>=6}}}.
As a consequence, no real number can be a solution of {{{x^2+6=0}}}.
The solutions are complex numbers, imaginary numbers, and you could find them, if you know about complex numbers.
What you probably were expected to do, is calculate the discriminant.
All quadratic equations can be represented by the general form
{{{ax^2+bx+c=0}}} with {{{a}}} different from zero.
In your case {{{a=1}}}, {{{b=0}}} and {{{c=6}}}
The discriminant is the expression {{{b^2-4*a*c}}}.
In your case {{{b^2-4*a*c=0^2-4(1)(6)=0-24=-24}}}.
The textbooks tell you that if the discriminant is negative there are no real solutions, but there are two imaginary solutions.
(They also tell you that if the discriminant is zero, there is one real solution, and if it is positive, there are two real solutions).