(b+4)(b-4)=6b

Solution for (b+4)(b-4)=6b equation:

(b+4)(b-4)=6b

We move all terms to the left:

(b+4)(b-4)-(6b)=0

We add all the numbers together, and all the variables

-6b+(b+4)(b-4)=0

We use the square of the difference formula

b^2-6b-16=0

a = 1; b = -6; c = -16;
Δ = b2-4ac
Δ = -62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*1}=\frac{-4}{2}
=-2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*1}=\frac{16}{2}
=8
$