Let $Q_k=(Q_{k,0},Q_{k,1})\in\mathbb R^2$ for $k=0,1,2$. And the same for $P_k\in\mathbb R^2$. Let $A=(A_1,A_2)$ be the Bezier curve with the points $Q_k$ and let $B=(B_1,B_2)$ be the Bezier curve with the points $P_k$.
Then Mathematica gives that $A_1(t)=B_1(t)$ is equivalent to (note that I am not assuming that $t\in[0,1]$ here, you will have to check that. Also, there are many conditions to avoid pathological cases such as the two Bezier curves being identical), where $\land$ is the logical and operator and $\lor$ is the logical or operator, $$\left(P_{0,0}+P_{2,0}+2 Q_{1,0}\neq 2 P_{1,0}+Q_{0,0}+Q_{2,0}\land
\left(t=-\frac{\sqrt{-2 P_{1,0} Q_{1,0}+P_{2,0} Q_{0,0}+P_{0,0}
\left(Q_{2,0}-P_{2,0}\right)+P_{1,0}^2+Q_{1,0}^2-Q_{0,0}
Q_{2,0}}+P_{0,0}-P_{1,0}-Q_{0,0}+Q_{1,0}}{-P_{0,0}+2 P_{1,0}-P_{2,0}+Q_{0,0}-2
Q_{1,0}+Q_{2,0}}\lor t=\frac{\sqrt{-2 P_{1,0} Q_{1,0}+P_{2,0} Q_{0,0}+P_{0,0}
\left(Q_{2,0}-P_{2,0}\right)+P_{1,0}^2+Q_{1,0}^2-Q_{0,0}
Q_{2,0}}-P_{0,0}+P_{1,0}+Q_{0,0}-Q_{1,0}}{-P_{0,0}+2 P_{1,0}-P_{2,0}+Q_{0,0}-2
Q_{1,0}+Q_{2,0}}\right)\right)\lor \left(P_{0,0}+P_{2,0}+2 Q_{1,0}=2
P_{1,0}+Q_{0,0}+Q_{2,0}\land P_{1,0}+Q_{2,0}\neq P_{2,0}+Q_{1,0}\land t=\frac{2
P_{1,0}-P_{2,0}-2 Q_{1,0}+Q_{2,0}}{2
\left(P_{1,0}-P_{2,0}-Q_{1,0}+Q_{2,0}\right)}\right)\lor
\left(P_{2,0}=Q_{2,0}\land P_{1,0}=Q_{1,0}\land P_{0,0}=Q_{0,0}\right)$$ and analogously $A_2(t)=B_2(t)$ is equivalent to $$\left(P_{0,1}+P_{2,1}+2 Q_{1,1}\neq 2 P_{1,1}+Q_{0,1}+Q_{2,1}\land
\left(t=-\frac{\sqrt{-2 P_{1,1} Q_{1,1}+P_{2,1} Q_{0,1}+P_{0,1}
\left(Q_{2,1}-P_{2,1}\right)+P_{1,1}^2+Q_{1,1}^2-Q_{0,1}
Q_{2,1}}+P_{0,1}-P_{1,1}-Q_{0,1}+Q_{1,1}}{-P_{0,1}+2 P_{1,1}-P_{2,1}+Q_{0,1}-2
Q_{1,1}+Q_{2,1}}\lor t=\frac{\sqrt{-2 P_{1,1} Q_{1,1}+P_{2,1} Q_{0,1}+P_{0,1}
\left(Q_{2,1}-P_{2,1}\right)+P_{1,1}^2+Q_{1,1}^2-Q_{0,1}
Q_{2,1}}-P_{0,1}+P_{1,1}+Q_{0,1}-Q_{1,1}}{-P_{0,1}+2 P_{1,1}-P_{2,1}+Q_{0,1}-2
Q_{1,1}+Q_{2,1}}\right)\right)\lor \left(P_{0,1}+P_{2,1}+2 Q_{1,1}=2
P_{1,1}+Q_{0,1}+Q_{2,1}\land P_{1,1}+Q_{2,1}\neq P_{2,1}+Q_{1,1}\land t=\frac{2
P_{1,1}-P_{2,1}-2 Q_{1,1}+Q_{2,1}}{2
\left(P_{1,1}-P_{2,1}-Q_{1,1}+Q_{2,1}\right)}\right)\lor
\left(P_{2,1}=Q_{2,1}\land P_{1,1}=Q_{1,1}\land P_{0,1}=Q_{0,1}\right).$$
We have $A(t)=B(t)\iff A_1(t)=B_1(t)\land A_2(t)=B_2(t)$.
Best Answer
It doesn't matter which one you substitute it in since it is a point of intersection, thus the $y$-coordinate of both graphs will be the same.
For example, let's try finding the coordinates of intersection where $x=4$:
$$f(x)=x^2-3x \Rightarrow f(4)=16-12=4$$ Now, let's try this with the other function: $$f(x)=3x^2-5x-24 \Rightarrow f(4)=48-20-24=4$$ Which is the same answer: $(4,4)$.
Here is a graph showing the intersection: