we can substitute y=2x in the quadratic equation to get:
x^2+(2x)^2=5
x^2+4x^2=5
5x^2=5
x^2=1
x=±1 so there are two solutions when x=-1 and 1, using y=2x we get the corresponding y values...
y(-1)=-2 and y(1)=2 so the two solutions are the two points:
(-1,-2) and (1,2)
