Hi im having problems in getting the exact value of the roots, for some reason it doesnt read -4+sqrt(13) as -0.3944487245 but rather as -4 + (-0.3944487245) which it turn shows the answer as -4.3944487245 which is wrong, what did i do wrong here???

Oh btw the reason that i did multiplication by 10^7 and got the floors then divided it back to 10^7 is because our professor wanted every calculation result at cut at exactly 7 decimal places and uses that value in any subsequent calculation

To calculate d in line 27, just do: b * b - 4 * a * c; it's not necessary to use Math.pow to square something.

Why are you calculating dFloor and dFinal? That's not part of the quadratic formula. What is all the multiplying and dividing by 10 to the power 7 for? It looks like you're making it way more complicated than necessary.

With the quadratic formula you need to take care when d is negative; there are no (real-valued) solutions in that case (because you can't take the square root of a negative number). And if d is equal to 0, there is only one solution instead of two.

Phil Maqui wrote:Oh btw the reason that i did multiplication by 10^7 and got the floors then divided it back to 10^7 is because our professor wanted every calculation result at cut at exactly 7 decimal places and uses that value in any subsequent calculation.

In which case I'd put it in a separate method (round(int)?), because it has nothing to do with calculating a quadratic and just confuses your readers.
Alternatively, you could just trim the excess when you print it out, viz:
System.out.printf("%.7f%n", r1);

Oh, and just to add to what Jesper said: (-1 * b) == -b.

I also tend to add brackets for clarity even if they're not strictly required, eg:
(-b + Math.sqrt(dFinal)) / (2 * a) rather than
-b + Math.sqrt(dFinal) / (2 * a) although I think most of yours are OK.

Winston

Isn't it funny how there's always time and money enough to do it WRONG?
Artlicles by Winston can be found here

To expand from what Jesper said: have a look at the Math class, and see what the sqrt() method returns from a negative input. You would have to test for that result specially.
If you have 0 for d, you would be adding ±√0, which is 0, so you would still get two results from a programmatic solution. Obviously both results should be the same!

Well the main reason why i just didn't get the roundoff value for each calculation is because in some cases it might change the values even by a little bit, and our prof WANTED the values to be cut and not to be rounded off at the 7th decimal place and use that value for subsequent calculations

But nonetheless i learned something and gained insight with all your comments, i figured that my mistake was with the grouping by brackets, since division takes priority over addition.