How to solve for $x$ in an equation such as "$2x^2 - 5x + 64\log(x) + 776 = 0$"? I'm trying to find the shortest distance between a point and a logarithmic line. So far I've applied the distance formula, found the derivative of the distance equation and made it equal to zero. That's how I got the equation in the title: $$2x^2 - 5x + 64\log(x) + 776 = 0.$$ From this point I haven't been able to figure out how to solve for $x$. So the question is, how to I solve for $x$ in this equation?
 A: first the derivation of this equation is greater than zero 
$y=2x^2-5x+64\log(x)+776$ so $\frac{dy}{dx}=4x-5+\frac{64}{x}=\frac{4x^2-5x+64}{x}$  in $4x^2-5x+64\geq 0$ since $\Delta \leq 0$ so always $\frac{dy}{dx}>0$
also $y(0^{+})=-\infty$ and there are some $y\geq 0$ so you have a one root and no more. (since $y$ is increasing and exist $a$ and $b$ such that $f(a)f(b)<0$و Intermediate value theorem)
R code
    y<-function(x){
    2*x^2-5*x+64*log(x)+776
      }
    uniroot(y,c(5.414414e-06,5.423423e-06))
   $root
   [1] 5.423423e-06

   $f.root
    [1] 0.01383579

   $iter
      [1] 0

   $init.it
   [1] NA

   $estim.prec
  [1] 9.009e-09

so you sure have a unique answer
A: I am pretty sure that it is impossible to solve for the equation algebraically. Let $f(x) = 2x^2 - 5x +64\log x+776$. We are trying to find when $f(x) = 0$. Note that the domain of $f$ is $(0,\infty)$ because of the logarithm( which I assume is the natural logarithm i.e. to the base $e$).
Going for it analytically, first we note that the derivative is $f' = 4x + \frac{64}{x} - 5 = \frac{4x^2 - 5x+64}{x}$ which  is strictly positive on $(0,\infty)$ since the numerator has discriminant negative , hence is always positive, while the denominator is always positive.
Also, from the fact that $f(x) \to -\infty$ as $x \to 0^+$ and $f(1) >0$, we see that any root is between $0$ and $1$.
We can actually do better : note that if $x = e^y$, where $y < 0$ since we now want $0<x<1$, then we have : $$2x^2 - 5x = 2(x^2 - 2.5x) = 2((x-1.25)^2) - \frac{25}{8}$$
Therefore, we see that on $[0,1]$ that this part is between $0$ and $-3$.
So if equality holds, then in particular, $64y + 776$ is between $0$ and $3$, so we get $\frac{-776}{64} \leq y \leq \frac{-773}{64}$, which after exponentiation gives $ 5.422\leq 10^6x \leq 5.682$. At least we get the number of significant zeros and the first significant non-zero digit, using at most two exponential operations.
