This is too long for a comment.
If we consider the equations
$$\begin{align}
y&=1-Ax^2 \\x&=1-By^2
\end{align}$$
eliminating $y$ to get the quartic in $x$ and then using the procedure given in this page, we have $$\Delta=A^4B^2(256 A^2 B^2-256 A^2 B-256 A B^2+288 A B-27)$$ $$P=-16 A^3 B^2 <0$$ $$Q=8 A^4 B^2 >0$$ $$\Delta_0=4 A^2 B (4 B-3)$$ $$D=-64 A^6 B^3<0$$ So, there will be four real roots if $\Delta >0$ and two real roots if $\Delta <0$.
That is to say that, for a given value of $A$, we should have only two real roots if $B$ is between the two roots $$B_1=\frac{A(9-8A)-\sqrt{A (4 A-3)^3}}{16 A(1-A)}$$ $$B_2=\frac{A(9-8A)+\sqrt{A (4 A-3)^3}}{16A(1-A)}$$ and four roots otherwise (this assumes $A\neq 0$).
For example, using $A=5$ and $B=0.9$ leads to two real roots while $A=5$ and $B=1.1$ leads to four real roots.
Looking at the particular case where $B=\frac 1A$ $$\Delta=-A \left(256 A^2-517 A+256\right)$$ which is positive if $$\frac{517-7 \sqrt{105}}{512}< A < \frac{517+7 \sqrt{105}}{512} $$ which represents a very narrow range.
Using $A=1.1$ leads to four real roots while $A=1.2$ leads to two real roots.
The problem seems to be quite sensitive to the values of the parameters.
Looking at the case where $A=1$, the problem simplifies a lot since $\Delta=B^2 (32 B-27)$. So, if $B > \frac{27}{32}$ four real roots and only two real roots otherwise.
Let us try for $B = \frac{26}{32}$
$${x= -1.67794}\,,{x= 0.338968 -0.150441 i}\,,{x= 0.338968 +0.150441
i}\,,{x= 1.}$$ while for $B = \frac{28}{32}$
$${x=-1.65597}\,,{x= 0.182018}\,,{x= 0.473952}\,,{x= 1.}$$
Edit
Interesting is the case where $B=A$; in such a case $\Delta=A^6 (4 A-3)^3 (4 A+1)$ and then four roots if $A >\frac 34$. If this is the case, the coordinates of the intersections are
$$\left(
\begin{array}{cc}
x & y \\
-\frac{\sqrt{4 A-3}-1}{2 A} & \frac{\sqrt{4 A-3}+1}{2 A} \\
\frac{\sqrt{4 A-3}+1}{2 A} & -\frac{\sqrt{4 A-3}-1}{2 A} \\
\frac{-\sqrt{4 A+1}-1}{2 A} & -\frac{\sqrt{4 A+1}+1}{2 A} \\
\frac{\sqrt{4 A+1}-1}{2 A} & \frac{\sqrt{4 A+1}-1}{2 A}
\end{array}
\right)$$
These points are along a circle centered at $\left(-\frac{1}{2 A},-\frac{1}{2 A}\right)$ with a radius equal to $R=\sqrt{\frac{1+4A}{2A^2}}$
Best Answer
In the same spirit as voldemort, let the two roots be $a$ and $b$. So, what you are given is $$a+b=\frac{55}{72}$$ $$a\times b=-\frac{25}{12}$$ But, you can use identities $$(a-b)^2=a^2+b^2-2a b=(a+b)^2-4a b$$ So, $$(a-b)^2=\Big(\frac{55}{72}\Big)^2+4\times\frac{25}{12}=\frac{46225}{5184}=\Big(\frac{215}{72}\Big)^2$$ For making life easier, suppose $a>b$; so now $$a+b=\frac{55}{72}$$ $$a-b=\frac{215}{72}$$