What is the correct way to calculate the discriminant in a quadratic equation?

[yungman]

I know for $ax^2+bx+c=0$,
$$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}$$

Using $x^2-3x-4=0$, we know it is equal to $(x+1)(x-4)=0$. So $x=-1$ or $x=4$.

but using the formula:

$$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}=\frac{3^+_-\sqrt{9+4}}{2}=\frac{3^+_-\sqrt{13}}{2}$$

I cannot get -1 and 4! What happened?

 

[Mentallic]

I know for $ax^2+bx+c=0$,
$$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}$$

Using $x^2-3x-4=0$, we know it is equal to $(x+1)(x-4)=0$. So $x=-1$ or $x=4$.

but using the formula:

$$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}=\frac{3^+_-\sqrt{9+4}}{2}=\frac{3^+_-\sqrt{13}}{2}$$

I cannot get -1 and 4! What happened?

How are you calculating the discriminant $\Delta = b^2-4ac$? Because you shouldn't be getting 13.
c = -4, not -1 which it seems that you've confused it for.

 

[yungman]

How are you calculating the discriminant $\Delta = b^2-4ac$? Because you shouldn't be getting 13.
c = -4, not -1 which it seems that you've confused it for.

b=-3, a=1 and c=-4, so $b^2-4ac$= 9+16=25.

Yes, I am missing the moon.

Thanks