This is the formula I worked out to use in my game to calculate player experience $x$ needed for level $y$:
C#: $x = (y * (y+1) / 2) * y^{1.5}$
or
$x = y \cdot (\frac{ y + 1 }{2}) \cdot y ^ {1.5}$
This I've been able to work out bits and pieces, but eventually keep getting stuck on having to solve for Y when only X is known. I am also limited on my math vocabulary in English (in addition to not paying enough attention at school previously) and am not familiar with all the learned notations, so most of the explanations that I find are way over my head.
I've worked out the logarithm to handle the exponent, the Cardano Method in this post, teied a few online equation solvers and eventually got to $2y = x\cdot\frac{7}{2}+x\cdot\frac{5}{2}$. However this doesn't result in anything sensible and I'm not sure what is wrong. The main problem is handling multiple use of $y$ on the right side, coupled with the exponent. Is there a somewhat simple way to explain how to work this out?
For example, when Y = 4, X = 80, so how would I get from 80 back to 4?
The equation write $$x=\frac{1}{2} y^{5/2} (y+1)$$ which corresponds to a polynomial of degree $7$ in $\sqrt y$ and this cannot be solved explicitly.
Numerical method being required, consider that you look for the zero of function $$f(y)= y^{5/2} (y+1)-2x$$ $$f'(y)=\frac{1}{2} y^{3/2} (7 y+5)$$ and use Newton method starting from a guess $y_0$ and obtain the iterates $$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$ A reasonable choice seems to be $y_0=(2x)^{2/7}$ (obtained neglecting the $1$ in the equation).
Let us try for your example $x=80$; the iterates would be $$\left( \begin{array}{cc} n & y_n \\ 0 & 4.263326233 \\ 1 & 4.018611856 \\ 2 & 4.000101025 \\ 3 & 4.000000003 \\ 4 & 4.000000000 \end{array} \right)$$
Edit
We can have a quite good estimation of the solution using rational approximation built from Padé approximants around $y0$. $$y=t\,\frac {15+77t+98t^2}{35+105t+98t^2}\qquad \text {where} \qquad t=(2x)^{2/7}$$ The table below shows some results $$\left( \begin{array}{ccc} x& \text{approximation} & \text{exact} \\ 50 & 3.46832 & 3.46724 \\ 100 & 4.28014 & 4.27936 \\ 150 & 4.83593 & 4.83530 \\ 200 & 5.27152 & 5.27097 \\ 250 & 5.63508 & 5.63459 \\ 300 & 5.94993 & 5.94949 \\ 350 & 6.22930 & 6.22889 \\ 400 & 6.48150 & 6.48111 \\ 450 & 6.71211 & 6.71175 \\ 500 & 6.92511 & 6.92477 \\ 550 & 7.12342 & 7.12309 \\ 600 & 7.30925 & 7.30894 \\ 650 & 7.48434 & 7.48404 \\ 700 & 7.65007 & 7.64978 \\ 750 & 7.80756 & 7.80728 \\ 800 & 7.95772 & 7.95745 \\ 850 & 8.10133 & 8.10107 \\ 900 & 8.23904 & 8.23878 \\ 950 & 8.37139 & 8.37114 \\ 1000 & 8.49885 & 8.49861 \end{array}\right)$$ This does not seem to be too bad !
Update
More empirical but simpler), based on a nonlinear regression for $0 \leq y \leq 10$, $$y=t \, \frac{a+b\,t}{1+b\, t}$$ leads to an almost perfect fit with $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.239090 & 0.000296 & \{0.238509,0.239672\} \\ b & 2.656562 & 0.001130 & \{2.654346,2.658779\} \\ \end{array}$$