Inspired by Theodore Jonathan

My solution is conceptually similar to Jake's.

Solving the equation for c, we find $c=f(x)=-\frac{168+2x^2}{x}$ . Using the derivative (or AM-GM), we find that $f(x)$ is increasing from 0 up to $\sqrt{84}$ and then decreasing. Thus the maximum for prime numbers x is attained at 7 or at 11. We find $c=f(7)=-38$ and $c=f(11)=\boxed{-\frac{410}{11}}\approx-37.27$ . (The other solution is $x=\frac{84}{11}$ in this case.)

Let the roots of the equation be $\alpha$ and $\beta$ . We know that $\alpha \beta = \frac{168}{2} = 84$ by and $\alpha + \beta = -\frac{c}{2}$ both by Vieta's formulas.

We know that primes must be positive integers, and as such we can work with AM-GM to find the minimum $\alpha + \beta$ (which implies the maximum $c$ ), which turns out to be $\alpha = \beta = \sqrt{84} \approx 9.17$ ; however, this is not a prime. So, we test primes in proximity to $\sqrt{84}$ , like $7$ or $11$ .

If $\alpha = 7$ , $\beta = \frac{84}{7} = 12$ and $\alpha + \beta = 19$ . However, if $\alpha = 11$ , $\beta = \frac{84}{11} = 7.6363\ldots$ and $\alpha + \beta = 18.6363\ldots$ .

Having found our prime, we can then calculate our $c$ :

$c = -2(\alpha + \beta) = -2 \times 18.6363\ldots \approx \boxed{-37.27}$

As one solution is prime, so c must be an integer. If c>0, then equation has no positive root. So, c must be negative.
If a and b are two roots then, a + b = - c/2 and ab = 84 = 2X2X3X7
If a is prime then the value of a can be 2, 3 or 7.
If a = 2, then b = 42, then c = - 88
If a = 3, then b = 28, then c = - 62
If a = 7, then b = 12, then c = - 38
So, - 38 is the maximum value of c.