(8b-1)(5b-1)=

Solution for (8b-1)(5b-1)= equation:

(8b-1)(5b-1)=

We move all terms to the left:

(8b-1)(5b-1)-()=0

We add all the numbers together, and all the variables

(8b-1)(5b-1)=0

We multiply parentheses ..

(+40b^2-8b-5b+1)=0

We get rid of parentheses

40b^2-8b-5b+1=0

We add all the numbers together, and all the variables

40b^2-13b+1=0

a = 40; b = -13; c = +1;
Δ = b2-4ac
Δ = -132-4·40·1
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*40}=\frac{10}{80}
=1/8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*40}=\frac{16}{80}
=1/5
$