I saw this question in a textbook:
Two taps fill a container in $\frac{75}{8}$ hours together. If one tap takes $10$ hours more than the other, how long does each tap take to fill up the container?
First approach:
Let slower tap to fill the container in $x$ hours and the faster one will take $x-10$ hours. Setting up the equation as,
$$ \frac{1}{x} + \frac{1}{x-10} = \frac{8}{75} $$
Solving this we get $x = 25$.
So taps take $15$ and $25$ hours to fill up the container.
This is the text book answer.
Second Approach:
Assume that the faster tap takes $y$ hours to fill the container, so the slower one will take $y+10$ hours. Setting up the equation, we get:
$$ \frac{1}{y} + \frac{1}{y+10} = \frac{8}{75} $$
Solving this we get, $y = \frac{25}{8} + \frac{5\sqrt{265}}{8}$, (ignoring the negative solution).
Hence the taps take $\frac{25}{8} + \frac{5\sqrt{265}}{8}$ and $\frac{105}{8} + \frac{5\sqrt{265}}{8}$ hours to fill up the container.
Question: Are both approaches correct? Are both solutions correct? Am I doing anything wrong?
Only the first approach is correct. There is something wrong with the second approach. The quadratic equation should be $$8y^2 - \color{red}{70}y - 750 = 0$$ but you seem to solve the quadratic equation $$8y^2 - \color{red}{50}y - 750 = 0$$ instead.