I think what I would, is first compute the first derivative:
$$y'=(x-1)(3x-(2a+1))$$
And then the second derivative:
$$y''=6x-2(a+2)$$
Okay, now let's look at the root of the first derivative:
$$x=\frac{2a+1}{3}$$
With this value for $x$, we require the second derivative to be positive:
$$6\left(\frac{2a+1}{3}\right)-2(a+2)>0$$
$$2a+1-a-2>0$$
$$a>1$$
This tells us that if the local minimum is going to occur anywhere other than that $(1,0)$, then we need $1<a$. And when we have $1<a$, the local minimum is locate at:
$$(x,y)=\left(\frac{2a+1}{3},\frac{4}{27}(1-a)^3\right)$$
Now, in order for this point to lie on the line $y=-4x$, we require:
$$\frac{4}{27}(1-a)^3=-4\left(\frac{2a+1}{3}\right)$$
$$(a-1)^3=9(2a+1)$$
$$(a+2)\left(a^2-5a-5\right)=0$$
Taking the only root such that $1<a$, we obtain:
$$a=\frac{5+3\sqrt{5}}{2}$$
And this agrees with the value you found. :D