Solving $4k^3 + 17k^2 - 228k -1116 = 0$ The equation given to me is $$4x^4 + 16x^3 - 17x^2 - 102x -45 = 0$$
I'm asked to find it's resolvent cubic which is not so difficult to find. But the problem is that the question further asks to find the solution of resolvent cubic.
I have found resolvent cubic using Ferrari's method. The resolvent cubic came out to be $$4k^3 + 17k^2 - 228k -1116 = 0$$
The trouble begins now when I'm trying to solve this cubic it's getting very exhaustive and long calculations. But since the question asks to find the solution of the cubic. 
Now can someone help me solve this cubic easily or there's no way out to this problem. I want to escape the tedious calculations while solving this cubic. 
Kindly help me if you can.

Ferrari's method: If $a$, $b$, $c$, $d$ are the roots of the quartic, form the equation with roots
$$a b + c d, a c + b d, a d + b c$$
The coefficients of this equation are symmetric in $a$, $b$, $c$, $d$ ( since we have $a b + c d$ and all its symmetric transforms).
After we solve the cubic, the quadric is solved using the equalities
$$(a+b+c+d)^2 - 4 ( a b + c d) - 4 ( a d + b c) = (a- b + c - d)^2$$
Now, with $u = a b + c d$, $v = a c + b d$, $w= a d + b c$ we have
$(u-v)(u-w)(v-w) = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$$
so the discriminant of the cubic resolvent equals the discriminant of the quadric.
If the quartic is
$$x^ 4 + s x^3 + p x^2 + q x + r$$
then the Ferrari cubic resolvent is
$$x^3- p x^2 + (q s - 4 r)x + (4 p r - r s^2 + 4 p r)$$
So, if the particular quartic is
$$4(x^4 + 4x^3 - \frac{17}{4}x^2 - \frac{51}{2}x -\frac{45}{4})$$
then the Ferrari resolvent is
$$x^3 + \frac{17}{4}x^2 - 57 x - 279$$
or
$$4 x^3 + 17 x^2 - 228 x - 1116$$
Calculating the discriminant of $4 x^3 + 17 x^2 - 228 x - 1116$  we get $0$. That means that the cubic has multiple roots. To find a multiple root, calculate the GCD of the cubic and its derivative
$$\textrm{gcd}(x^3 + \frac{17}{4}x^2 - 57 x - 279,  3 x^2 + \frac{17}{2}x - 57) = x+6$$
Now, divide the cubic $x^3 + \frac{17}{4}x^2 - 57 x - 279$ by $(x+6)^2$ and find $x-\frac{31}{4}$.
