Solution below assumes knowledge of the quadratic formula for the equation

where solution
x=

and where the expression

is called the discriminant.
Given system of linear equations:
y=3 ..................................(1)
y=ax^2+b.........................(2)
[ note: b here is the constant term, and is "c" in the quadratic formula]
Need to find a and b so that system has exactly two solutions, i.e. two distinct solutions, not one, not zero.
Substitute (1) in (20
3=ax^2+b
rearrange
ax^2+b-3=0 ........................(3)
Solve by the quadratic formula:
x=(-0 ± √ (0^2-4a(b-3))/2a
For x to have two distinct solutions, the discriminant must be greater than zero.
The discriminant is the expression inside the square-root sign, namely
0^2-4a(b-3)
=-4a(b-3)
case A: a=-2,b=2, discriminant = -4(-2)(2-3)=-8 <0 (no real solution)
case B: a=-2,b=4, discriminant = -4(-2)(4-3)=+8 >0 (2 real distinct solutions)
case C: a= 2,b=4, discriminant = -4(2)(4-3)=-8 <0 (no real solution)
case D: a= 4,b=3, discriminant = -4(4)(3-3)= 0 =0 (two coincident solutions)
So from the previous calculations, the only case that gives two distinct real solutions is case B, which gives a discriminant >0.