How do we find the angular, radial nodes and the quantum numbers when the radial probability distribution curve's equation is given? I am having serious conceptual issues with the Schrödinger equations. Please note that I only have basic school level (11th grade) knowledge of these equations and we are expected to know very little about it. But we are given these type of questions so I guess I will be able to understand things within the boundary of this question.  

$$\psi_r = k_1 * e^{-r/k_2} * (r^2 - 5k_3r + 6k_3^2)$$ 

Solving the equation we get $r = 3 k_3 , 2 k_3$ 

It is also given that $k_3 =1$ . So $r = 3, 2$


Now, since there are 2 nodes therefore, $n = 3$ . Does the equation $\psi_r$ represents only radial nodes or all nodes?
What is the azimuthal quantum number?
How many angular nodes are present?
What is the angular momentum of the given orbital?
Is this equation of an atom or an orbital?


This equation was given as an MCQ question and we were asked to find the quantum states of the given orbital on the basis of the equation. But I guess we must have conceptual insight too to answer the question so asked it here.
For a given orbital, there are two types of modes: radial nodes and angular nodes.


The total number of nodes is $n-1$
Number of angular nodes is $l$
Number of radial nodes is $n-l-1$


In this question: If the given function represents the radial part of the wave function of an atomic orbital, then we get the number of radial nodes: $n-l-1=2$. i.e.  $n-l=3$:

We have several possibilities: 


$n=3$, $l=0$: then we have the orbital $\ce{3s}$
$n=4$, $l=1$: then we have the orbital $\ce{4p}$
$n=5$, $l=2$: then we have the orbital $\ce{5d}$


and so on.