demonstrate how to do the quadratic formula. I need more than

one person to give it some variation. So if you're interested...

To begin, make sure you have set up the equation properly...

x= (-(b) +/- Sqrt( b^2 -4 (a)(c)))/(2*a) //read as X equals negative B plus or minus the square root of B squared minus 4 times A times C ALL over 2 times A

Now, you input your variables A, B, and C.

Remember that at the beginning it is Negative B, as in -1*B

We then solve for the discriminant (that is the expression under the radical).

Multiply A times C, then take that product and multiply it by negative four.

Then we square b, and add the two expressions.

Get the square root of the number.

Multiply your A times 2.

Next, we must separate the equation into two separate equations.

Then we reduce.

We are done.

//Ok that was the base now we have to go ever the dependent stuff

minus

add

divide

multiply

square root

square

i

You will notice that our discriminant is negative. It is impossible to get the square root of a negative number. We must make it positive, and add an i to the end of it.

The square root of this discriminant is not an integer (it has a decimal) so it is appropriate, in most cases, to leave it as the square root of the discriminant.

Hopefully, you remembered to include the i in the final solution, because if you didn't then your answer is wrong.

Now, if the discriminant is positive you will get two real numbers, if it is negative you will get two complex numbers, and if it is 0 then you will get 1 repeating real number.

Would you like to hear more about 'i?'

Okay.

I is an imaginary number; i equals the square root of -1.

So you can deduce that i^2 //i squared

equals 1, this is important later on.

You cannot get the square-root of a negative and expect to come up with anything sensible, or so you have been told thus far.

In reality, you

*can*, take the square root of a negative number, using i.

YOU MUST ALWAYS DO THE i-PART FIRST!!!

You deal with i just as if it were x, except for the fact that x^2 is just x^2, but i^2 is