Been using computers too long?

I think I've been using computers way too long. I am having the darnedest
time trying to remember basic algebra. I am drawing a complete blank with:

Given the formula: ax^2 + bx + c = y (standard parabola)

and I know the values of a b c and y, how to solve for x?

Thanks,
Jeff

 

google would help. For example:

http://www.1728.com/quadratc.htm

 

Here's how I'll do it:

1) Transpose c and divide by a:
X^2 + (b/a)x = (y-c)/a

2) Add b^2/(2a)^2 to both sides:
X^2 + (b/a)x + b^2/4a^2 = (y-c)/a + b^2/4a^2

3) Take square root of both sides
(x + (b/2a) = [b^2 + (y-c)a]^1/2

4) Solve for x and simplify:

X = (-b +-[b^2 + (y-c)a]^1/2)/2a

HTH.

Leo

 

Thanks, I DID Google, just used incorrect terminolgy......

and once I saw this I recognized it as something that once was etched into
my brain, I guess the sands of time have started eroding things I once
thought untouchable....

Egads, my Dad told me to watch for these days.... ;-)

Jeff

 

Thanks, Leo

And that's exactly how I remember doing it, oh so many years ago.......

Thanks for shining the light through this foggy old brain....

Jeff

 

oops, you'v lost something in step 3:

(x + (b/2a) = (1/2a) * [b^2 + 4(y-c)a]^1/2

X = (-b +-[b^2 + 4(y-c)a]^1/2)/2a

 

You're welcome. Thanks to Bob for
for the corrections. -Leo

 

for^2 ?

 

Hi Doug,
Thanks for the attachment. After trying the other suggestions and getting
invalid results, I plugged in the formula from you, er MathCad, and it works
as expected.

I think that MathCad is in my future ;-)

Now to figure out how to know which one to use....

Thanks again,
Jeff