Answer
See the explanation
Work Step by Step
$F=\frac{-K}{x^2}+\frac{0.012K}{(239-x)^2}$,
$F=0$,
$\frac{-K}{x^2}+\frac{0.012K}{(239-x)^2}=0$,
$\frac{-K(239-x)^2+0.012K(x^2)}{x^2(239-x)^2}=0$,
$-K(239-x)^2+0.012K(x^2)=0$,
$-K(57,121-478x+x^2)+0.012K(x^2)=0$,
$-57,121K+478Kx-Kx^2+0.012Kx^2=0$,
$-57,121K+478Kx-0.988Kx^2=0$,
$0.988Kx^2-468Kx+57121K=0$, After multiplying both side by -1.
Solving for the trinomial using quadratic formula for the quadratic equation of $ax^2+bx+c, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, $0.988Kx^2-468Kx+57121K$, $a=0.988K$, $b=-478K$, $c=57,121K$.
Therefore, $x=\frac{478K\pm\sqrt{(-478K)^2-4(0.988K)(57,121K)}}{2(0.988K)}=\frac{478K\pm52.36K}{1.976}$.
Therefore, $x=268.4$ or $x=215.4$
