(7b-8)(b-5)=0

Solution for (7b-8)(b-5)=0 equation:

(7b-8)(b-5)=0

We multiply parentheses ..

(+7b^2-35b-8b+40)=0

We get rid of parentheses

7b^2-35b-8b+40=0

We add all the numbers together, and all the variables

7b^2-43b+40=0

a = 7; b = -43; c = +40;
Δ = b2-4ac
Δ = -432-4·7·40
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-27}{2*7}=\frac{16}{14}
=1+1/7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+27}{2*7}=\frac{70}{14}
=5
$