Silage cattle calculation

According to the plan, the cooperative secured 210 tons of silage for the winter. However, it then bought 10 heads of cattle, so it was necessary to reduce the amount of silage by half a ton per head. How many tons of silage did the cooperative initially plan for one head of cattle?

Final Answer:

x =  3.5 t

Step-by-step explanation:

$$\begin{gathered} 210=nx \\ 210=(n+10)(x-0.5) \\ 210=(n+10)(210/n-0.5) \\ 210n=(n+10)(210-0.5n) \\ 0.5n^2+5n-2100=0 \\ a=0.5;b=5;c=-2100 \\ D=b^2-4ac=5^2-4\cdot 0.5\cdot (-2100)=4225 \\ D>0 \\ {n}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-5\pm \sqrt{4225}}{1}=-5\pm 65\sqrt{1} \\ {n}_{1,2}=-5\pm 65 \\ {n}_1=60 \\ {n}_2=-70 \\ n>0 \\ n=n_1=60 \\ x=210/n=210/60=3.5\text{t} \end{gathered}$$

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