WakeUp wrote: Let $r,s,t$ be the roots of the equation $x(x-2)(3x-7)=2$. Show that $r,s,t$ are real and positive and determine $\arctan r+\arctan s +\arctan t$. Put $f(x) = x(x - 2)(3x - 7)-2 = 3x^3-13x^2+14x-2$. Then $f(0)=-2$, $f(1)=2$, so there is a root between $0$ and $1$. $f(2)=-2$, so there is another root between $1$ and $2$. $f(3)=4$, so the third root is between $2$ and $3$. $f(x)=0$ has three roots, so they are all real and positive. We have: $tan(a + b + c) = \frac{tan a+tanb+tanc-tana*tanb*tanc}{1-(tana*tanb + tanb*tanc+tanc*tana)}$. So putting: $a=\arctan r$, $b=\arctan s$, $c=\arctan t$ we have: $tan(a+b+c)=\frac{(r+s+t)-rst}{1-(rs+st+tr}=\frac{13/3-2/3}{114/3}=-1$. So $a+b+c=-\pi/4 + k\pi$. But we know that each of $r, s, t$ is real and positive, so $a+b+c$ lies in the range $0$ to $3\pi/2$. Hence $a+b+c = 3\pi/4$.