If the flight path of a cricket ball is given by: $$y = \frac{1}{3}x - \frac{1}{60}x^2$$ And a fielder standing originally at $(10, 0)$ catches the ball when it is $1.5$ units above the ground, to find how far he ran you would simply sub $1.5$ in as $y$ and solve for $x$ from a quadratic and take into consideration where he started at, correct? I've tried this but the number I get is not rational. It does not factorise, so I filled into the quadratic formula to get: $$10 + \sqrt{10}$$ or $$10 - \sqrt{10}$$
It just seems odd to me, so I'm checking here that I'm doing it correctly.