Solving two curves Let two curves be:-
S1: $x^2/9+y^2/4=1$
S2: $y^2=2x$
Now, on solving these two ,by substituting $y^2$ as $2x$ from S2 in S1, I get two values viz. $x=3/2$ and $x=-6$.
But we can clearly see from the graph that the curves intersect at two distinct points in the first and fourth quadrant. So, should not the quadratic equation hence formed in $x$ rather be a whole square viz $(x-3/2)^2$ ?

Two conics usually have four intersection points (by elimination of one variable you obtain a quartic equation). In this particular case, they are
$$\left(\frac32,3\right),\left(\frac32,-3\right),\left(-6,i\sqrt{12}\right),\left(-6,-i\sqrt{12}\right).$$
