Ladle size that halves wine in two draws

A keg holds $10$ gallons of wine. On each of two days a full ladle is taken out and the keg is topped back up with water, with the contents mixed before the second draw. Afterwards the keg is half wine, half water. The ladle's capacity is $a-b\sqrt c$ gallons where $a=bc$. Find $a+b+c$.

17. Each draw keeps a fraction $1-\tfrac{x}{10}$ of the wine, so $10\left(1-\tfrac{x}{10}\right)^2=5$, giving $1-\tfrac x{10}=\tfrac1{\sqrt2}$ and $x=10-5\sqrt2$. Then $a=10,b=5,c=2$ and $a+b+c=17$.