# Need another confirmation

1. May 5, 2005

### abia ubong

while working on the quadratic formula in school ,i came across another formula,and wanted to know if its been derived since i got this from the quadratic formula.the formula is as follows
[-b^3-2abc +or- sqrt(b^6-12a^2b^2c^2-16a^3c^3)]/(2ab^2+4a^2c).
please i want to know if it works for all.

2. May 5, 2005

### dextercioby

$$\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-b^{3}-2abc\pm\sqrt{b^{6}-12a^{2}b^{2}c^{2}-16a^{3}c^{3}}}{2ab^{2}+4a^{2}c}$$

...?There's only one way to find out:CROSS MULTIPLY...

Daniel.

3. May 5, 2005

### shyboy

looks like the multiplication of both the nominator and the denominator by
$$b^{2}+2ac}$$

4. May 5, 2005

### inha

regardless of if it works or not the form on the right is highly useless.

5. May 6, 2005

### abia ubong

that was a hard thing to say inha, how useless is it ,i just came across it and felt like letting the forum know how correct it is and u call it useless.
thae was a hard thing 2 say

6. May 6, 2005

### dextercioby

I'm sorry not to share your disspointment,but it's nothing more than a waste of ink and paper (or server space)...

Daniel.