(R-4)(r-4)(r+8)=0

Solution for (R-4)(r-4)(r+8)=0 equation:

(-4)(R-4)(R+8)=0

We multiply parentheses ..

(-4R+16)(R+8)=0

We multiply parentheses ..

(-4R^2-32R+16R+128)=0

We get rid of parentheses

-4R^2-32R+16R+128=0

We add all the numbers together, and all the variables

-4R^2-16R+128=0

a = -4; b = -16; c = +128;
Δ = b2-4ac
Δ = -162-4·(-4)·128
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-48}{2*-4}=\frac{-32}{-8}
=+4
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+48}{2*-4}=\frac{64}{-8}
=-8
$