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

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

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

We multiply parentheses ..

(+4b^2+8b-16b-32)=0

We get rid of parentheses

4b^2+8b-16b-32=0

We add all the numbers together, and all the variables

4b^2-8b-32=0

a = 4; b = -8; c = -32;
Δ = b2-4ac
Δ = -82-4·4·(-32)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-24}{2*4}=\frac{-16}{8}
=-2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+24}{2*4}=\frac{32}{8}
=4
$