Not shure what I should do by 'quadratic'
$\displaystyle \displaystyle{\frac{500+x}{17,5}}=\sqrt{x²+80x}$
$\displaystyle \displaystyle{\frac{500²+x²+1000x}{17,5²}}=x²+80x$
$\displaystyle x²+80x-\displaystyle{\frac{500²}{17,5²}-\frac{x²}{17,5²}-\frac{1000}{17,5²}x}=0$
$\displaystyle (17,5²-1)x²+(17,5²*80-1000)x-500²=0$
$\displaystyle 305,25x²+23500x-250000=0$
$\displaystyle x²+\displaystyle{\frac{23500}{305,25}}x=250000$
I guess....
$\displaystyle y(x)=\frac{250000*305,25}{23500}+Ce^{-\frac{23500}{305,25}x}$
But then I need some help to be able to solve x for y=0 ...
The last quoted line is OK.
A quadratic equation
$\displaystyle a x^2+bx +c = 0$ has the solution $\displaystyle x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
you know: a = 305.25, b = 23500, c = -250000
so you can calculate the value of x. (Btw the negative solution isn't very plausible with your question)