(b-2)(b-4)=A

Solution for (b-2)(b-4)=A equation:

(b-2)(b-4)=

We move all terms to the left:

(b-2)(b-4)-()=0

We add all the numbers together, and all the variables

(b-2)(b-4)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

b^2-6b+8=0

a = 1; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·1·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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