I'm getting a little confused with the quadrativ formula, i used to know it, but its been a while!

i'm getting -b+/- square root (b-4ac) / 2* c

the part im getting confused on is the 2 * c, what if my original equation was x^2 - 10x - 3 (where -3 would be c)

do i add the minus sign into the equation which would give me 2 * -3 = -6 and then divide the whole equation by -6 or do i forget the minus sign and change it to positive 6?

2. The quadratic formula is $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

2a on the bottom NOT 2c.

Hello
Originally Posted by entrepreneurforum.co.uk
I'm getting a little confused with the quadrativ formula, i used to know it, but its been a while!

i'm getting -b+/- square root (b-4ac) / 2* c

the part im getting confused on is the 2 * c, what if my original equation was x^2 - 10x - 3 (where -3 would be c)

do i add the minus sign into the equation which would give me 2 * -3 = -6 and then divide the whole equation by -6 or do i forget the minus sign and change it to positive 6?
You're asking two things here: the first is about the formula itself, and the second is how to handle minus signs. So:

1 The formula you want is

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

2 Handling minus signs. Remember the rules for multiplying and dividing positive ( $+^{ve}$) and negative ( $-^{ve}$) numbers:

$+^{ve}\times +^{ve}=+^{ve}$

$+^{ve}\times -^{ve}=-^{ve}$

$-^{ve}\times +^{ve}=-^{ve}$

$-^{ve}\times -^{ve}=+^{ve}$

So, do you just forget the minus signs and change them to positive? Definitely not!

$x^2 - 10x - 3$ gives $a=1$, $b=-10$ and $c=-3$

Plugging these values into the formula, you get:

$x=\frac{-(-10) \pm \sqrt{(-10)^2-4(1)(-3)}}{(2)(1)}$

(Where I've written two numbers in brackets next to one another - like $(1)(-3)$, there's a $\times$ sign in between; so that means $(1) \times (-3)$.)

So $x = \frac{10 \pm \sqrt{100+12}}{2}$

Can you see where all the minus signs have gone, using the rules above?

I hope that helps to clear one or two problems up.