Given the function f(x)=3x+5a / x^2-a^2 find the value of a for which f'(12) = 0

The function, as presented here, is ambiguous in terms of what's being deivded by what.  For the sake of example, I will assume that you meant

           3x+5a
 f(x)=
          x^2-a^2

You are saying that the derivative of this function is 0 when x=12.  Let's differentiate f(x) with respect to x and then let x = 12:

             (x^2-a^2)(3) -(3x+5a)(2x)
f '(x) = = 0 when x = 12
                [x^2-a^2]^2

(144-a^2)(3) - (36+5a)(24)
 =  0
               [   ]^2

Simplifying,

(144-a^2) - 8(36+5a) = 0

144 - a^2 - 288 - 40a = 0

This can be rewritten as a quadratic in standard form:

-a^2 - 40a - 144 = 0, or a^2 + 40a + 144 = 0.

Solve for a by completing the square:

a^2 + 40a + 20^2 - 20^2 + 144 = 0
        (a+20)^2    = 400 - 144 = 156

        Then a+20 = sqrt[6(26)] = sqrt[6(2)(13)] = 4(3)(13)= 2sqrt(39)

         Finally, a = -20 plus or minus 2sqrt(39)

You must check both answers by subst. into the original equation.  Only if the result(s) is(are) true is your solution (value of a) correct.

Ithink the answer is discovery verbs because they are quite similar to the verb 'discovering'