This one is fun! However, that particular formula is not my favorite, as it doesn't behave nicely when a is small. The art here is deciding what to special case

This one is fun! However, that particular formula is not my favorite, as it doesn't behave nicely when a is small. The art here is deciding what to special case. The way I approach this[1] is to choose the sign of the ± to be the same as -b to compute the first root x1. Then the second root is c/(a * x1). There are a few other checks in the code for things overflowing, but that basically gives you very accurate answers across the range. This is explained a bit in a good Math Exchange post[2]. Jim Blinn's "How to solve a quadratic equation" also makes good reading. Wait til you get to cubics and quartics. [1]: https://docs.rs/kurbo/latest/src/kurbo/common.rs.html#116-160 https://docs.rs/kurbo/latest/src/kurbo/common.rs.html#116-16... [2]: https://math.stackexchange.com/questions/866331/numerically-stable-algorithm-for-solving-the-quadratic-equation-when-a-is-very https://math.stackexchange.com/questions/866331/numerically-...