Is the "quadratic()" function being properly calculated?

[d4v3y5c0n3s]

I recently asked another question about the "quadratic()" function, but I have another one about the implementation. This function obviously lets one use the quadratic formula, which is x = (-b +- sqrt(b^2 - 4 a c))/2a. Please see this Khan Academy if my quadratic formula is unclear. This is almost reflected in the code, but instead of "2a," it appears to divide the discriminant by just "2." Why is this? I'm fairly new to computer graphics math, so I am unsure if this is intentional or not.

Here is the code from Corange that I'm asking about, with the parts I'm curious about followed by a comment. `bool quadratic(float a, float b, float c, float t0, float t1) {

float descrim = bb - 4a*c;

if (descrim < 0) {

return false;

} else {

float d = sqrtf(descrim);
float q = (b < 0) ? (-b - d) / 2.0 : (-b + d) / 2.0;//  Should "2.0" be changed to "2.0*a"?

*t0 = q / a;
*t1 = c / q;

return true;

}

}`

[orangeduck]

Interesting - I can't actually remember where I got this formulation from.

I can see that in the end t0 = q / a so the final value of the first solution is getting divided by a - just one step after it has been divided by 2. However I can't remember how or why the second solution t1 can be derived as being c / q in this case.

[d4v3y5c0n3s]

@orangeduck Yea, I know t0 & t1 are used to calculate the "time" struct parameter in cphysics, but I don't know how this works. Can you point me to any resources that you used when developing this? I want to understand the math here if possible.

[orangeduck]

After a bit more poking around it looks like I am using what is essentially this formulation:

https://math.stackexchange.com/a/2007723

Where q = x_1 * a. In my code t0 and t1 correspond to the two different solutions which on this page are denoted x_1 and x_2.

[d4v3y5c0n3s]

@orangeduck thank you for the article, now it makes a bit more sense what's going on in the code here. Should I close the issue, or leave it open for others to see?

[orangeduck]

Thanks, I will close it!