Solve for y in terms of x then solve for some (x, y) coordinates to draw its graph
Using quadratic formula
rewrite the equation first, so that it appears as follows
#y^2+sqrt3xy+2x^2-10=0#
Let #a=1# and #b=sqrt3x# and #c=2x^2-10#
#y=(-b+-sqrt(b^2-4ac))/(2a)#
#y=(-sqrt3x+-sqrt((sqrt3x)^2-4(1)(2x^2-10)))/(2*1)#
#y=(-sqrt3x+-sqrt(40-5x^2))/2#
Compute y values for some x arbitrary values
#y=(-sqrt3x+sqrt(40-5x^2))/2#
#" "x" " " " " " "y#
#" "0" " " " "+3.16#
#+0.5" " " " "2.679#
#+0.75" " " " "2.399#
#+1" " " " "2.092#
#+1.25" " " " "1.754#
#+1.5" " " " "1.3819#
#+1.75" " " " ".9687#
#+2" " " " "0.504#
#+2.25" " " " "-0.0323#
#+2.5" " " " "-0.68604#
#+sqrt3" " " " "1#
#" "x" " " " " " "y#
#-0.5" " " " "3.54548#
#-0.75" " " " "3.698#
#-1" " " " "3.824#
#-1.25" " " " "3.919#
#-1.5" " " " "3.9799#
#-1.75" " " " "3.9998#
#-2" " " " "3.968#
#-2.25" " " " "3.864#
#-2.5" " " " "3.644#
#-sqrt3" " " " "4#
Kindly see the graph of #y^2+sqrt3xy+2x^2-10=0#
graph{y^2+sqrt3xy+2x^2-10=0[-sqrt8,sqrt8,-5,5]}
God bless you...
