Find the point of equilibrium on two formulas Sorry I don't really know how to express my issue.
I am playing an online game and I am trying to find which weapon is the best for my character.
The formula is as follows :
weaponDamage = baseDamage * (1 + ( power / 100 )) + runes

(The power is a %-based damage, while the rune is fixed bonus damage)
What I would like to know is when the first weapon is better than the second weapon, depending on the power and runes.
First weapon :

*

*100 baseDamage

*Hits once

Second weapon :

*

*12 baseDamage

*Hits four times

For instance, with 100 power and 0 runes, then the first weapon is better (200 vs 96).
But with 0 power and 100 runes, the second one is better (200 vs 448).
What I would like to know is the tipping point at which one weapon is better than the other.
So I wrote the following formula where X is the power and Y the runes :
100 * (1 + (x / 100)) + y = (12 * (1 + (x / 100)) + y) * 4

which I simplified to (not sure it's correct but it's a try)
y = (13 * (x + 100)) / 75

The issue I am facing now is : how should I read this information ?
From my understanding, I read it like this :
If I have 100 power, the result is approx. 34.6, so I need more than 35 runes for the second weapon to be better than the first one

Is that a correct assumption ? If not, then how should I read it ?
Thank you in advance !
 A: So, your equation relating the two weapons is the correct way to find the point where they cross over in damage, and you've correctly rearranged to give $y$ as a function of $x$. Your conclusion about how to compare the two weapons for a fixed power of 100 as the runes vary is also correct. Great question and good work!
I'll add in a couple of additional thoughts which you might find helpful.
You have correctly understood how to find the points where the weapons crossover in damage dealt for a fixed power as the runes vary, expressed as $$y(x) = \frac{13}{75}(x +100).$$ If you wanted look at a situation where are the runes stay the same but the power varies, you could rearrange to give $x(y)$. I'm not sure if this is relevant to your game, but you might find it instructive as an exercise.
When I looked at your problem I found the concept of a function to be quite helpful. The first thing I did was to introduce a damage function based on the first expression that you gave at the top, $$D(n,b,p,r) := n(b(1+\frac{p}{100})+r).$$ What this notation means is, if you give me a number of hits, base damage, power and runes, I can plug them into this formula to give the weapon damage $D$. I find it's helpful to name my variables after what they refer to so I know what's going on.
From this I then defined a damage function for each weapon that you gave me,$$W_1(p,r):= D(1,100,p,r) = 100(1+\frac{p}{100})+r = 100 +p +r,$$ and similarly I defined $W_2(p,r):=D(4,12,p,r)$ (the subscripts 1 and 2 don't mean anything special, it's just to refer to the first and second weapon). Then to find where the two weapons deal equal damage, I wrote down the equation $$W_1(p,r) = W_2(p,r),$$ which led me to the same condition that you found, except using $p$ and $r$ as the names for my variables instead of $x$ and $y$. The result is the same as what you did, I thought it might be helpful to introduce you to some new notation.
Just a minor point; you tagged the question as calculus. Calculus is when you introduce integration and differentiation; your question involves only algebra, and not calculus.
Welcome to stack exchange. I hope you have fun learning maths, analysing your video games is definitely a great way to do it!
