some polish

docs/chapters/extremities/content.en-GB.md

@@ -87,7 +87,7 @@ Back in the 16<sup>th</sup> century, before Bézier curves were a thing, and eve
   \end{aligned}
 \]
 
-We can see that the easier formula only has two constants, rather than four, and only two expressions involving `t`, rather than three: this makes things considerably easier to solve because it lets us use [regular calculus](http://www.wolframalpha.com/input/?i=t^3+%2B+pt+%2B+q) to find the values that satisfy the equasion.
+We can see that the easier formula only has two constants, rather than four, and only two expressions involving `t`, rather than three: this makes things considerably easier to solve because it lets us use [regular calculus](https://www.wolframalpha.com/input/?i=t^3+%2B+pt+%2B+q) to find the values that satisfy the equasion.
 
 Now, there is one small hitch: as a cubic function, the solutions may be [complex numbers](https://en.wikipedia.org/wiki/Complex_number) rather than plain numbers... And Cardona realised this, centuries befor complex numbers were a well-understood and established part of number theory. His interpretation of them was "these numbers are impossible but that's okay because they disappear again in later steps", allowing him to not think about them too much, but we have it even easier: as we're trying to find the roots for display purposes, we don't even _care_ about complex numbers: we're going to simplify Cardano's approach just that tiny bit further by throwing away any solution that's not a plain number.
 
@@ -196,7 +196,7 @@ And this is where thing stop, because we _cannot_ find the roots for polynomials
 
 That's a fancy term for saying "rather than trying to find exact answers by manipulating symbols, find approximate answers by describing the underlying process as a combination of steps, each of which _can_ be assigned a number via symbolic manipulation". For example, trying to mathematically compute how much water fits in a completely crazy three dimensional shape is very hard, even if it got you the perfect, precise answer. A much easier approach, which would be less perfect but still entirely useful, would be to just grab a buck and start filling the shape until it was full: just count the number of buckets of water you used. And if we want a more precise answer, we can use smaller buckets.
 
-So that's what we're going to do here, too: we're going to treat the problem as a sequence of steps, and the smaller we can make each step, the closer we'll get to that "perfect, precise" answer. And as it turns out, there is a really nice numerical root-finding algorithm, called the [Newton-Raphson](http://en.wikipedia.org/wiki/Newton-Raphson) root finding method (yes, after *[that](https://en.wikipedia.org/wiki/Isaac_Newton)* Newton), which we can make use of. The Newton-Raphson approach consists of taking our impossible-to-solve function `f(x)`, picking some intial value `x` (literally any value will do), and calculating `f(x)`. We can think of that value as the "height" of the function at `x`. If that height is zero, we're done, we have found a root. If it isn't, we calculate the tangent line at `f(x)` and calculate at which `x` value _its_ height is zero (which we've already seen is very easy). That will give us a new `x` and we repeat the process until we find a root.
+So that's what we're going to do here, too: we're going to treat the problem as a sequence of steps, and the smaller we can make each step, the closer we'll get to that "perfect, precise" answer. And as it turns out, there is a really nice numerical root-finding algorithm, called the [Newton-Raphson](https://en.wikipedia.org/wiki/Newton-Raphson) root finding method (yes, after *[that](https://en.wikipedia.org/wiki/Isaac_Newton)* Newton), which we can make use of. The Newton-Raphson approach consists of taking our impossible-to-solve function `f(x)`, picking some intial value `x` (literally any value will do), and calculating `f(x)`. We can think of that value as the "height" of the function at `x`. If that height is zero, we're done, we have found a root. If it isn't, we calculate the tangent line at `f(x)` and calculate at which `x` value _its_ height is zero (which we've already seen is very easy). That will give us a new `x` and we repeat the process until we find a root.
 
 Mathematically, this means that for some `x`, at step `n=1`, we perform the following calculation until `f<sub>y</sub>(x)` is zero, so that the next `t` is the same as the one we already have: