(5b-5)(2b-3)=0

Solution for (5b-5)(2b-3)=0 equation:

(5b-5)(2b-3)=0

We multiply parentheses ..

(+10b^2-15b-10b+15)=0

We get rid of parentheses

10b^2-15b-10b+15=0

We add all the numbers together, and all the variables

10b^2-25b+15=0

a = 10; b = -25; c = +15;
Δ = b2-4ac
Δ = -252-4·10·15
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5}{2*10}=\frac{20}{20}
=1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5}{2*10}=\frac{30}{20}
=1+1/2
$