Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$ Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$If $k,w,f$ are relatively prime positive integers,then find the value of $k+w+f.$

The given equation is $\log_8(kx^2+wx+f)=2$ i.e. $kx^2+wx+f=64$
Since $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,so
$\frac{k}{9}+\frac{w}{3}+f=64.............(1)$
$225k-15w+f=64............................(2)$
I need to find $k,w,f$ from these equations,but i need three equations to find three variables.I am stuck here.

Hint $gcd(k,w,f)=1$ as they are relatively prime. Or thats a quadratic so product of roots is $-5=\frac{f-64}{k}$ and sum of roots is $\frac{44}{3}=-\frac{w}{f}$ can you get all in w and solve it.
