solving for variable: is it possible to solve R^2 +10R = 2000 be R = something?

[Bronn]

is it possible to make this equation:

R^2 +10R = 2000

be R = something?

i can't figure it out if there is.

I was working on a word problem and got the equation down to this trying to find the rate of something. But this is where Im stuck. I do know that the R^2 +10R = 2000 is correct as i know the answer and it fits if you punch it in.

 

[ksdhart2]

Yes, definitely, although there will be two possible values of R. Let's rearrange the equation a bit, by subtracting 2000 from both sides. That leaves R² + 10R - 2000 = 0. That's a quadratic equation that you can factorhttp://www.purplemath.com/modules/factquad.htm using your preferred method.

 

[Bronn]

of course doh!

Im new to quadratics so Im not trained to look for them enough yet

thanks


(-10 +/- √100+8000)/2

(-10+/- √8100)/2

-5+/- 45

-5+45 = 40

thats it

 

[mmm4444bot]

-5+45 = 40

thats it

That's half of "it". :cool:

 

[Bronn]

That's half of "it". :cool:

well its a rate word problem finding the speed of a train, and unfortunately this is a regular non-interdimensional train confined to only positive kmph so I left that answer out.