(b-2)(b-1)=

Solution for (b-2)(b-1)= equation:

(b-2)(b-1)=

We move all terms to the left:

(b-2)(b-1)-()=0

We add all the numbers together, and all the variables

(b-2)(b-1)=0

We multiply parentheses ..

(+b^2-1b-2b+2)=0

We get rid of parentheses

b^2-1b-2b+2=0

We add all the numbers together, and all the variables

b^2-3b+2=0

a = 1; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*1}=\frac{2}{2}
=1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*1}=\frac{4}{2}
=2
$