Calculus problem (already know the answer, just need help explaining)

This is from a problem I got wrong, I just want to see if anyone can walk me through where I went wrong?

Find the extrema of G(x)= -x^3+4x^2+16x-4

I did the differentiating to get -3x^2+8x+16

then factored it to get -(x-4)(3x+4)

I then got for my numbers -4 and -3/4

I plug the 4 into the original equation to get 60, which makes (4,60) my maxima

But how do I find the minimum? The answer is (-4/3, -428/27) so I guess I screwed up with the 3/4 as a number but can someone help me here?

factor out -1
-(3x^2-8x-16)
-(3x+4)(x-4)=0
x=-4/3 and x=4
plug in x=-4/3.
G(-4/3)=(-4/3)^3+4(-4/3)^2+16(-4/3) - 4
G(-4/3)=-64/27+64/9-64/3-4=

I then got for my numbers -4 and -3/4

solve -3x^2+8x+16=0
3x^2-8x-16=0
The numbers should be 4 and -4/3 ----------------------------->


-3x^2+8x+16=0
3x^2-8x-16=0
3x^2-12x+4x-16=0
3x(x-4)+4(x-4)=0
(x-4)(3x+4)=0
x=4 or x=-4/3

I guess I screwed up with the 3/4 as a number but can someone help me here? -- yes