(Wolframalpha) this shouldnt give x=50? This is my problem:
I need x or y for the triangle area that forms between the vertical axis(y) and the function y=100+2x where the area is equal to 2500.
so I used for condition to the linear function:
knowing that the triangle area in this case should be like:
x*y/2=area, so:
x-100*y/2=2500

x-100*y=5000

y=5100/x

and then:
5100/x=100+2x

5100=100+2x*x

5000=2x^2

sqrt(2500)=x

50=x

the weird thing is that works for any area, and gives me the correct result for what I'm looking for, wich is x=50 and y=f(50)=200, if the area is calculated as is shown in the condition: 200-100*50/2=2500 !
5100/x=100+2x [http://www.wolframalpha.com/input/?i=5100%2Fx%3D100%2B2x][1]

it outputs x=-5 (5+sqrt(127)) and x=5 (-5+sqrt(127))
how can I get the same results on wolframalhpa ?
thank you ! (:
 A: You made an error when multiplying $x$ on both sides. 
You had $$\frac{5100}{x} = 100 + 2x$$
In order to remove the $x$ from the denominator you correctly decided to multiply by $x$ on both sides. However, when you do this you should get $$x \frac{5100}{x} = x(100 + 2x)$$
And multiplying $x$ through the equation becomes
$$ 5100 = 100x + 2x^2$$
and now I'm sure you can solve it! 
EDIT: To address the edit in your question, 
be careful when setting up the area of this triangle because it is actually 
$$\frac{1}{2} x(y-100)$$ 
since the bottom part of our triangle is located at $y = 100$. 

And to solve this area to be 2500, we would plug in $$ \frac{1}{2} x(y-100) = 2500$$
And we know that $y = 100 + 2x$, so plugging that in for $y$ gives us $$ \frac{1}{2}x(100 + 2x - 100) = 2500$$
and canceling out the $100$s and multiplying through by the $x$ and $\frac{1}{2}$ gives $$ x^2 = 2500$$ or $x = 50$. 
A: You start with the equation $\frac{5100}{x} = 100 + 2x.$ If the left side and the right side are equal, then I can do the same to both sides and they'll still be equal. Let's multiply both sides by $x$. I get $5100 = 100x + 2x^2.$ Bringing all of the terms over to one side, we get $2x^2 + 100x - 5100 = 0.$ Next, notice that there is a common factor: $2(x^2 + 50x - 2550) = 0.$ Finally, we use the quadratic formula where $a = 1,$ $b = 50$ and $c = -2550$. We have:
$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \, ,$$
$$x = \frac{-50 \pm \sqrt{50^2-4\times 1 \times (-2550)}}{2 \times 1} \, , $$
$$x = \frac{-50 \pm \sqrt{12700}}{2} \, , $$
$$x = \frac{-50 \pm 10\sqrt{127}}{2} \, , $$
$$x = -25 \pm 5\sqrt{127} \, .$$
It seems that the website was correct. Notice that $-5(5\pm \sqrt{127}) = -25 \mp 5\sqrt{127}.$ If you don't see why $\sqrt{12700} = 10\sqrt{127}$ then notice that $12700 = 2^2 \times 5^2 \times 127$, where $127$ is prime and so:
$$\sqrt{12700} = \sqrt{2^2 \times 5^2 \times 127} = \sqrt{2^2} \times \sqrt{5^2} \times \sqrt{127} = 2 \times 5 \times \sqrt{127} = 10\sqrt{127} \, . $$ 
