As written there is an infinite number of "solutions" in the form of #(x,y)#pairs for which the equation is valid, including: #(0,6), (1,5), (-2,2)#, etc.

I will assume you wanted the solution for the x intercept, which is equivalent to the solution to

#color(white)("XXXX")##0 = -x^2+6#

#color(white)("XXXX")##color(white)("XXXX")#(if this is not the case, re-post your question with clarification)

If #-x^2+6 = 0#

then, equivalently,

#color(white)("XXXX")##x^2-6 = 0#

#color(white)("XXXX")##x^2 = 6#

#color(white)("XXXX")##(x+0)^2 = 6##color(white)("XXXX")#would technically "complete the square... but to what point?

#color(white)("XXXX")##(x+0) = +-sqrt(6)##color(white)("XXXX")#carrying on as if it were necessary

#color(white)("XXXX")##x = +-sqrt6#