<s>[INST] Steps to get Inverse of Pentagonal I have solved http://projecteuler.net/problem=44 by getting the inverse equation from Wikipedia http://en.wikipedia.org/wiki/Pentagonal_number:
Pentagonal:
$f(n) = \frac{n(3n - 1)}{2}$
Inverse Pentagonal:
$n = \frac{\sqrt{24f(n) + 1}+1}{6}$
am interested in the steps from Pentagonal equation (quadratic?) to the Inverse.
I note that it is similar to What is the inverse of $f(n)=\frac{n^2+n}{2}$?   and I've tried the same strategy:
$f(n) = \frac{n(3n - 1)}{2}$
*6 + 1 on each side
$6f(n) + 1 = 9n^2 -3n + 1$  
but this isn't correct because I want:
$6f(n) + 1 = 9n^2 -6n + 1$
to give:
$(3n-1)^2$ on the right hand side
 [/INST] To imitate the procedure used in the solution of the other problem, starting from
$$2f(n)=3n^2-n,$$
multiply both sides by $12$. We get
$$24f(n)=36n^2-12n.$$
Note that $36n^2-12n=(6n-1)^2-1$. The rest is easy. We get $(6n-1)^2=24f(n)+1$, then $6n-1=\sqrt{24f(n)+1}$, then $6n=\sqrt{24f(n)+1}+1$.
Another way to solve the same problem is to write our equation as 
$$3n^2-n-2f(n)=0,$$
and use the Quadratic Formula.
Remark: Look at the quadratic equation $ax^2+bx+c=0$, where $a\ne 0$. Multiply both sides by $4a$. We get the equivalent equation
$$4a^2x^2+4abx+4ac=0.$$
Note that $4a^2x^2+4abx=(2ax+b)^2-b^2.$
So quickly we arrive at the equation
$$(2ax+b)^2=b^2-4ac.$$
From this we conclude that 
$$2ax+b=\pm\sqrt{b^2-4ac},$$
and then straightforward algebra yields
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$$
the important Quadratic Formula.
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